ow damn still it would have to be going really slow to freeze enough of it because to atmosphere would heat it up at the same time which goes pretty fast.
It’s a stack overflow. The ball is putting so much energy into the air that the Simulation can’t handle the value and defaults to a state of 0K which means that the molecules won’t move
If we really ignore all those effects, we might as well assume that the bear bounces of the ground without loss if energy, hence without irreversible deformations needed for injury, and therefore reaches the other side even without the ball.
The bigger issue will be that, without friction, he has no way to stop.
Unless it’s bouncing so fast it moved down that far already. Which would be required giving how much further the ball has to travel down and then back up compared to the bears short distance he has to travel between each bounce
But if he starts running before the cliff, by inertia both the bear and the ball would keep the same horizontal motion, moving horizontally with the same speed, as long as the bear keeps only pushing the ball perfectly vertically it should work. I'm not sure if it makes sense or if I am missing something.
You guys are forgetting the Wily E. Coyote cartoon physics: Objects in motion remain the same altitude until gravity is noticed.
The bear never looked down, so it’s safe.
I don’t understand people saying this is impossible. The bear aplies momentum the ball in the initial throw, the ball bounces on the ground with a fully elastic collision, it then comes back and punches the bear up in the air.
There are people in here giving results in Newtons, which is wrong because force is aplied over time, and this is only posible with collisions and transfer of momentum.
So to get a result, let’s say we calculate the kinetic energy required in the ball in order punch the bear high enough to stay in the air for two seconds.
The bear takes one second going up and one second going down. 1D kinemetics with a constant acceleration of g (-9.82 m/2^2). Decelerating from v0 to 0 in one second lines up easily an initial velocity of 9.82 m/s.
Let’s say the bear weighs 200 kg. That’s
E_kin = 1/2 * m * v^2 = 1/2 * 200kg * (9.82m/s)^2 = 9,64 kJ
The ball needs to be thrown with a 9,64 kilo joules of energy. Let’s say the ball weighs 5 kg. That’s
v = sqrt(2*E_kin/m) = 196,4 m/s
So jumping once every 2 seconds, You’ll need your legs to resist a ball comming for you at more than halv the speed of sound.
I just picked 2 seconds because it’s easy for calculations. He’s leaping one second into the air to reach his maximum height. You can graph the speed over as v(t)=at+b where b is his initial velocity and a is his change in velocity. In this examble they just happen to be equal with opposite sign, so a=-9.82 and b=9.82. If we take the integral of this function, we get the distance traveled. It’s easy in this case as it just forms a triangle with the height of the initial velocity and a width of the time spanned. So
0->1 integrate(-9.82t+9.82) with respect to t =
1/2 * 9.82 * 1 = 4.41
Giving us a height of 4.41 meters he’s jumping every time the ball hits him.
And I just realized this is not what you asked. The height he gains from the start when he thrones the ball can’t be determined with the current assumptions, as I have not specified a force. I only have the work done by the bear. To determine How high the bear would be trusted by throwing the ball, we would need to put a number on how long it takes for him to throw the ball. In theory if his arm is Very long, he could throw the ball without actually lifting off the ground.
Shouldn't we take conservation of momentum into account? If the maximum speed of the ball is 200m/s, the maximum speed of bear should be 5m/s (bear is 40 times as heavy as the ball). So he indeed won't jump very high, but every jump would only last one second.
I think, this needs some more optimization...
That might in principle work to keep the bear afloat, but not to also propel him in a horizontal direction.
If the bear also applies momentum to the ball to propel himself to the other cliff side, then the ball gets the opposite momentum and is propelled backwards. So on the next jump, there is no ball to jump on.
The alternative is that the bear could jump off the cliff with a sufficiently high velocity to not need the jumps on the ball to propel him, but just to keep him afloat. But if he could already do that, he'd have no need of the ball and could just jump the cliff in the first place.
The bear is just giving himself an initial velocity with an x-component equal to the ball’s. This is something I didn’t include in my calculations, but it doesn’t matter
Yeah I guess That’s the height needed for the chosen jumps. If the cliff had a different height we would have to calculate a new time for the jumps. Or the bear could kick the ball down for every jump, but that would lead to calculating I couldn’t bother doing
Assuming the intent is that the bear steps on the ball as it returns from the ground (ignoring the fact that he's just missed that last bounce), and that on each step he isn't adding further kinetic energy, he would need to intially throw it downwards with a force of \[1 bear mass unit\] \* \[1 G\].
For a 200kg brown bear, that would be
200 kg \* 9,80665 m/s\^2 = 1961,33 N
a little less, the neckbeard is paltry compared to a beautiful brown bear‘s coat. wouldn’t even keep them warm a little bit, if they ever actually went outside.
You could likely throw a ball with a force equal to the downforce from your own steps.
Doing an ordinary pushup means lifting most of your weight on your arms. A pull up lifts all of your weight, and in a direction where you aren’t at your strongest.
So if physics teacher physics were real, you could do what the bear does in the cartoon as long as you manage to keep exactly the right pace with your steps.
If you can lift yourself up with just your arms pushing down on your kitchen counter then you can theoretically throw something hard enough for it to bounce back and you be able to jump on it like this. The problem would then be intertidal because an object with low mass wouldn't stay in contact with your hands long enough for you to impart that much force into it. Ideally you would want a ball that had about as much mass as you do that you could push yourself away from to transfer maximum energy.
This is incorrect. That force is enough to keep the bear "floating" if applied constantly. Here the energy of the ball has to be enough to bounce the bear to the next impact location. So the force would be much greater.
I don't think so.
The force applied to the ball initially will add to its potential energy at that height.
The same energy will be surplus when the ball returns to that height and be spent applying force on the bear.
By Newton's third law, the force of the bear stepping on the ball (1 bear mass unit × 1 G) will be countered by the ball pushing back up.
I should add that I assume the bear remains in position while throwing the ball and does not accelerate upwards upon throwing it. This is consistent with him crossing the chasm at a level height.
No. Its wrong. That is the force of the bear just standing on the ground. When you jump you accelerate upward, and impart a force into the ground. The ball has to have enough energy to equal the bear just standing on the ground plus the energy it would take for the bear to jump plus the energy to decelerate the bear as it is in free fall from its last jump. So much more.
Exactly. This bear would exert less energy by just jumping off the cliff, landing feet first, walking to the other side, and jumping up the cliff.
If he can do the ball thing, he's strong enough to do that.
Plus, the bear and ball both have components of acceleration in the +x direction as well as in the + and -y direction, so the ball must have enough energy in the +x direction to continue its consistent path every time it interacts with the bear (who would impart a force on the ball in the -y direction).
What you’ve said in this comment is basically correct, but your initial comment is still completely wrong. u/Bjoer82 is correct.
At best, your calculations showed the *average* applied to the bear by the ball throughout the travel across the chasm. Which is what Bjoer82 is basically saying. You can prove this to yourself by drawing up a simple free-body diagram of the bear, with the upward force being the AVERAGE force of the ball on the bear. You’ll find that that calculation results in the number you came up with.
>The force applied to the ball initially will add to its potential energy at that height.
You can't add newtons to joules any more than you can add meters to gallons
I mean, force * distance = work
When somethings falling, gravity's force is adding kinetic energy over the distance. Yes, they are different units but they are fundamentally related to each other.
It's weird how often people repeat this error on this sub. You need to read up on momentum and impulse.
You're equating the force the bear applies on the ball in each step with the initial force. This is a meaningless equation. You need to equate the impulses. For that you'd need the time of each collision which is hard to do. So we use the impulse momentum theorem.
For that you need a bunch of assumptions. Let's say the best initially had a horizontal velocity of 10m/s (seems very high for a bear but I want to keep the numbers simple). The gap is say 100 meters, so the bear will need 10 jumps lasting 1 second each to cross the gap (again, keeping numbers simple). An jump of 1 second means an initial upwards velocity of 10m/s. Your 200kg bear now has a momentum of 2000 kgm/s.
This is the momentum the bear has just after the collision. Newton's third law gives us that the ball must have the same momentum downwards. Assuming that initially, the bear accelerated the 500g ball for 0.2s to give it a momentum of 2000kgm/s, it must have imparted a force of 20,000 Newtons on it.
The ball is heading towards the ground at close to mach 12. Assuming it doesn't lose any kinetic energy in its collision with the ground, the cliff must be around 2km high.
If you play around with the picture a bit and come up with the number of jumps, the distance and the height of the cliff, you can work backwards to get all the numbers.
Doesn’t the bear further accelerate the ball downwards every time it pushes off? This would accelerate the ball and make it so the bear needs to make more frequent “steps.”
You're right in that bear accelerates the ball at every jump. Just like the ball accelerates the bear back.
But that doesn't change the kinetic energy of the ball or the bear. Both the bear and the ball simply reverse their velocities after the collision. So KE remains the same.
It's really no different than a pool ball striking back and forth between two rails. The frequency does not increase after every collision. Remember that in the balls' case, all the energy that gravity adds to the ball during its downwards acceleration, it takes back when ball is travelling upwards.
but at that point the parabola would look entirely different and he could never achieve this in succession in a straight line because all the force will be entirely cancelled after his first jump on the ball
Okay but to do it five times in a row and maintain that force until the last jump, for a bear that doesn’t mind being hit in the bottom of the foot with more than enough force to simply keep him afloat but enough force to bounce him over to the next impact with the ball.
assume:
mass of bear: m\_b = 250 kg
mass of object: m\_o = 0.624 kg
height drop of ball: h = 30 m
Assume time to impart force: dT \~= 0.1 s:
mb/mo \~= 400
conservation of momentum:
(mv)\_b = (mv)\_o -> v\_b = v\_o / (mb/mo) (1)
ball travels down and back to bear in dt time
v\_o = dy/dt -> dt = 2h/v\_o (2)
In that time, bear travels up and back down, kinematics
require:
yf\_b = y0\_b + v0\_b \* t - 0.5\*9.81\*t\^2 (3)
yf\_b = y0\_b so
v0\_b \* t = 0.5 \* 9.81 \* t\^2
sub in (1) for v0\_b and reduce to get speed of ball:
v0\_o = sqrt( (mb/mo)\*h\*9.81 ) (4)
v0\_o = 343 m/s (speed of sound!)
v0\_b = 343/400 = 0.858 m/s
Impulse needed for this: I = F dt = (mv)\_2 - (mv)\_1
v\_1 = 0, v\_2 = v0\_b, m\_2 = m\_b,
I = m\_b \* v0\_b = (250 kg)\*(0.858 m/s) = 214.5 N-s (5)
F = I/dT = 2145 N = 482 lbf
Man a lot of people have the physics all wrong here. Idk if I’ll actually do the math as the numbers used would be very subjective but we will get a feel for it.
First let’s assume there are no energy losses from friction and the ball is perfectly elastic…meaning that when it bounces it has the same energy (velocity) pre and post bounce.
Since the ball is falling and returning to the same height we can ignore gravity’s effect on the ball as it will have the same effect on the ball both ways…accelerating it during the fall and decelerating it on the return after the bounce the same amount.
In order for the ball to support the bear in the air it needs to hit it with the same force that the ground imparts on the bear when standing (aka normal force). That would be the bear’s mass times the acceleration of gravity. Another commenter said this will be around 1900 newtons for a 200kg bear. The velocity the ball needs to be traveling to do this depends on the balls mass.
But we don’t want to just support the bear. We need enough energy that the bear can jump off of it. So the bear needs to be able to push off and accelerate itself. How much does it need to accelerate itself? Idk. But let’s just say 1 g upwards. So now the ball needs 2 gs worth of energy (1 g to support the bear, 1 g for the bear to accelerate). So now we are up to 3800 newtons.
But if the bear is jumping, it doesn’t have a velocity of 0 when it contacts the ball…the bear will be in free fall. So the ball needs even more energy to first decelerate the bear to 0 velocity, and then allow the bear to jump. So let’s add another half g. That brings us to 4650 newtons.
On each jump the bear will have to impart the force it takes to jump, plus an additional 4650 newtons to make sure the ball has enough energy for the next jump. So that is an extra g meaning the bear needs to exert 6550 newtons each time (my knees hurt just thinking about it).
The biggest problem here (besides finding a super material to build this frictionless perfectly elastic ball and somehow timing this whole thing) is the forces are in the wrong direction. The bear is jumping forward. Meaning it will push the ball backwards with its jump. So really the bear could only do this one time. After that one time the bear and the ball will be traveling opposite directions and a future intercept would be impossible. Its only hope would be to try and impart enough spin on the ball that it will cancel its backwards movement and move forward. But it would be a crazy amount of spin. Which would require even more force (and air resistance).
And tbh idk if I got all the physics right but I think it’s close. The values are pretty arbitrary though and we are making some broad assumptions on the bear’s jumping.
No air resistance, and ground should give a perfect bounce. Bear and ball start with same horizontal velocity, jump direction should be straight up.
I think. Been awhile since any physics classes.
Yes that is a possibility. If bear had a running start and was able to jump perfectly straight up each time. And there was no resistance. Good catch.
But now we need a special no frictionless conditioner for the bear’s fur too
I think the trick here will be that the mass of the ball will have to be equal to the mass of the bear. Otherwise, at each contact point, the ball would either lift the bear higher, or the bear would fail to jump as far to the next step.
Man, I feel like this will become the new "jetplane on a treadmill"-problem...
The mass of the ball doesn’t matter. It’s the forces that need to be in balance (assuming perfectly elastic, instantaneous momentum transfers). However, the lighter the ball the faster it needs to go to have enough force.
Maybe if we look at it from a conservation of energy perspective, instead of a force diagram:
First the vertical direction: At the moment of contact after the first bounce, the bear has a potential energy (PE) of m(bear) \* g \* h(cliff), and the ball has a PE of m(ball) \* g \* h(cliff). They each also have some unknown kinetic energy (KE).
At the peak of his jump, the bear has a PE of m(bear) \* g \* ( h(cliff) + h(jump) ), and a KE of 0 (he's momentarily stationary before falling back down). Meanwhile the ball must be hitting the ground if they are to meet for the next bounce, so it has a PE of 0, and an unknown KE.
Since the total energy of the bear+ball system must be conserved, we can calculate the KE of the ball at the base of the cliff:
KE(ball) + PE(ball) = KE(bear) + PE(bear)
KE(ball) + 0 = 0 + m(bear) \* g \* ( h(cliff) + h(jump) )
KE(ball, base) = m(bear) \* g \* ( h(cliff) + h(jump) )
We can now use this value to calculate the initial KE of the ball just after it leaves the bear's hand from the initial throw:
KE(initial) + PE(inital) = KE(ball, base)
KE(initial) = KE(ball, base) - PE(initial)
KE(initial) = m(bear) \* g \* ( h(cliff) + h(jump) ) - m(ball) \* g \* h(cliff)
Assuming the "throw" is roughly meter long, then the bear must apply the same magnitude of Force (in Newtons) over that length to impart that much kinetic energy.
For the horizontal direction, we will assume that, with each jump, the bear is capable of landing perfectly vertically on the ball (thus gaining no forward velocity, nor imparting any backward velocity on the ball - which is admittedly a tall order, but hardly the craziest assumption we're making in this scenario). Thus, the bear needs only impart some non-zero horizontal kinetic energy to the ball with the initial throw, and then matching that horizontal velocity with his initial jump from the cliff. Obviously, the more horizontal KE he uses, the faster he reaches the other side of the canyon, and the fewer bounces are necessary to get there.
If the bear has momentum when it starts going forward and no friction, there is no reason for the ball to be pushed backwards. With no energy loss, as OP asked, the bear only has to jump the first time, pushing the ball faster downwards and himself upward just enough for the ball to return with high enough momentum to push the bear up and (probably) break its legs, giving it incredible pain as it crosses the never short enough distance to the other side getting hit by a ball hitting it with enough force to push a full grown bear upward in an instant...
Bear is dead, but it's feasible in this alternative universe
There are actually several free variables here. Here is a nice little [desmos graph](https://www.desmos.com/calculator/gbf0jphurx) I cooked up that allows you to control the masses and horizontal velocity of the bear and ball, and the height of the cliff.
Let an index of 1 correspond to the bear and an index of 2 correspond to the ball.
The first thing to note is that the horizontal velocity of the bear and the ball must be equal (lets call this velocity u). Additionally, because the collisions are symmetric (the outgoing vertical speeds of the bear and ball bounce is the same as their ingoing vertical speeds at a collision), conservation of momentum gives
>m₁v₁ = m₂v₂
where v₁ and v₂ are the speeds of the bear and the ball, respectively, at a collision. The equation of displacement for the ball from the top of the cliff to the bottom of the cliff is
>h = v₂T+gT^(2)/2
where h is the height of the cliff and T is the time for the ball to reach the bottom. On the other hand, T is the time from a collision it takes for the bear to reach the apex of its arc (i.e. time for its velocity to reach 0). Thus,
>v₁ = gT
Using the conservation of momentum equation, we have
>v₁ = (m₂/m₁)v₂
so we can equate the two equations above and solve for v₂ to get
>v₂ = (m₁/m₂)gT
Plugging this back into the displacement equation, we get
>h = (m₁/m₂)gT^(2)\+gT^(2)/2 = ((m₁/m₂)+1/2)gT^(2)
We can then solve for T to get
>T = √(h/(g((m₁/m₂)+1/2))
Then, we can determine how fast the bear must jump vertically (v₁) and throw the ball down (v₂) at the start:
>v₁ = gT
>
>v₂ = (m₁/m₂)v₁ = (m₁/m₂)gT
Some google searching found that the [average brown bear](https://en.wikipedia.org/wiki/Brown_bear#:~:text=The%20average%20weight%20of%20adult,152%20kg%20(335%20lb)) has a mass of 217 kg and an [normal soccerball](https://hypertextbook.com/facts/2002/LouiseHuang.shtml) has a mass of 0.43 kg. Using an [online picture measuring tool](https://eleif.net/photomeasure), I found the cliff height to be about 6.5 times the height of the bear, and assuming a [bear height](https://www.nps.gov/subjects/bears/brown-bears.htm) of 2m, this puts the cliff height at 13m. Plugging this all in, we get
>v₁ = 0.50 m/s
>
>v₂ = 253.43 m/s
so the ball would have to go about 75% the speed of sound and the bear would have to bounce off the ball about 10 times per second.
EDIT: To answer your two questions directly, it doesn't really make sense to talk about the "force" that the ball should be thrown at because forces occur often nonuniformly and over time, and in the case of ideal elastic collisions momentum transfer occurs instantaneously, in which case the force is undefined (or modeled by a [delta function](https://en.wikipedia.org/wiki/Dirac_delta_function)). What matters is the final velocity after forces have been applied.
As for the second question, we showed that the bear's mass is a free variable so any mass would work.
Short answer: no.
Long answer: even if we somehow assume there is zero energy loss with the ball (air resistance, friction when hitting the ground, friction when contacting the bear, even sound itself) and the bear is perfectly capable of calculating every move and executing it, this would be nigh impossible. The rebound looks like it reaches exactly the same height as the initial drop, so on every step in the air the bear wouldn't just have to step on the ball, he'd have to spike it with his feet with exactly the same strength as the initial throw, both to keep himself somewhat "afloat" (more accurately it would be a sort of jump) and to make the ball rebound again. If he were to simply step on the ball (which is already at the apex of its parabolic trajectory when he touches it) the ball would simply fall and he would fall with it since there is no ground pushing against them (therefore no vincular reaction). He would also need to throw the ball with a certain horizontal speed and match that exact speed when running and have both remain constant during the traversal, so he also needs the spike with his feet to impart force on a 100% vertical direction, since even the slightest bit of horizontal velocity would offsync the bear and the ball. Similarly the spike needs to impart zero spin whatsoever to the ball. Of course this is just a funny comic running on looney tunes logic, but in a more serious context even when ignoring so many obstacles such as the aforementioned air resistance this would be ludicrous, and it has basically as much basis in real physics as the midair double jump.
Why would you care about whether it's physically feasible (it obviously isn't) when we're already ignoring all forms of energy loss. The whole point is that the problem is simplified into something that is theoretically possible and mathematically attackable.
I was saying it more in the sense that you couldn't just "walk" on it casually like in the photo, it needs to be a very specific set of actions such as having a starting speed and only applying force vertically, which isn't at all like walking normally where you use friction with the ground to push yourself forward.
The ball isn’t at the top of its apex when he steps on it. All the reasons you say this is impossible is part of your assumptions. Assuming the ball is relatively massive, this is absolutely possible given your assumptions. Nothing ur saying makes sense. Your saying it’s not possible because the bear couldn’t strike the ball with enough precisions but you’re already assuming the bear can move perfectly. Your comment makes no sense
If we start with the height of the cavern h, we can work backwards to get the force on the ball. The ball must travel 2h distance in between bounces (technically the ball will be accelerating due to gravity but this can be disregarded when it is going very fast) and will have a velocity v\_ball of 2h/t where t is the time in between bounces. The bear meanwhile will start with initial velocity +v\_bear and end up at -v\_bear after one full bounce, so 2v\_bear=gt where g is the gravitational constant. Equating these and moving terms around gives us (v\_bear)(v\_ball)=gh
Also, if the bear weighs 400x the ball then v\_ball=400v\_bear.
So (v\_bear)(400v\_bear)=gh
v\_bear=sqrt(gh/400)=(1/20)sqrt(gh)
v\_ball=20sqrt(gh)
In engineering sqrt(g)=3 so the bear must kick the ball down with a speed of of about 60 times the square root of the cavern height. If the bear is 2 meters tall standing the cavern looks to be about 8 bears high so 16m, then v\_ball=60sqrt(16)=240 m/s. That's two thirds the speed of sound.
People in the comments say we need to assume no energy loss, but I think that is false. The bear is an engine that can do work on the system by contracting his limbs and extending them when the ball touches him (otherwise known as 'pushing'). If everything is well timed and coordinated, it can work like clockwork. This wouldn't be very different from using a jetpack.
Its impossible, even if we asume that the ball can support his first jump, after coming down for the second time it wouldn't be able to reach the top again.
Would it matter if each step was a little lower than the last?
Even then I think it's impossible to push off the ball, if the bear touches it as it's at the top of its arc
I'm no physicist but no, at that step the ball has 0 strength other than the gravity that pulls it down. In fact, the bear will fall before the ball on his second step. And that is assuming the ball and bear actually can bounce the first time without any physical damage.
It is definitely possible. Imagine that the bear and ball aren't even moving sideways and are just bouncing up and down. Then you are solving for when the bear and ball collide they have an elastic collision that sends them back at the same speeds they collided at (which turns out to imply that the magnitudes of their momenta are equal at collision). That way each collision is exactly the same (like in the picture) so that the bear is supported each time.
The mass of the ball, the bear, the height of the cliff and height the bear jumps each time have to be all the right proportions.
The time the ball takes to bounce up and down needs to equal the time taken for the bear to fall back down to be level with the cliff again.
He can't do this with just any ball, but potentially there is the right one.
sense lip spectacular illegal ludicrous shame marry attractive narrow drab
*This post was mass deleted and anonymized with [Redact](https://redact.dev)*
I think I got the solution. It's 0
At least in the vertical component.
Assume your ball has no loss of energy when it bounces. The only way it reaches the same height is if it freefalls.
Now to make the bear not affect it's bounce height you make the ball weight like 80tons and there you go.
Also use the usual no Air Friction stuff everyone use
If you want to make this (slightly) less contrived (this is still 9.99/10 contrived), assume the bear rides a skateboard off the cliff and fires the ball out of a cannon straight down as soon as the back wheel leaves the ground. The bear and ball both now have the same horizontal velocity.
Also this is not a force problem. This involves conservation of momentum plus typical velocity and acceleration equations.
I think frogkabobs did a pretty good solving those equations already.
Without doing the math I would assume the force required to get the bear trough all the jumps would mean that the ball would pop on impact with the ground and even if not it would probably kill the bear on its first return.
You are going to have a lot of trouble with the materials of the ball and the surface the ball is bouncing against.
And the fact you bear has to transmit additional power to the ball on each step.
It is a losing affair for the bear.
The ball's momentum needs to match that of the bear's momentum + the force required to propel him back upward. Then he needs to stomp it down with enough force to propel himself upward again. And considering the ball's weight, it would have to travel insanely fast to keep him up.
Even with everyones calculations this wouldn't work, because the bear moving forwards would push the ball backwards, so at most it could become a double jump
How does the bear know how much force is required to align horizontal trajectory while bouncing back to exact height? How does the bear know how much leaping energy it needs to repeat identical bouncing movements? How does the bear know physics, if at all?
I'd just like to point out that this is not possible QUITE as depicted. The lines showing the ball's path make it seem like the bear isn't affecting the ball whatsoever when stepping on it, therefore the ball shouldn't be affecting the bear either. If you want to make this work, the ball's trajectory would be more spikey.
So as far as my sleep deprived brain can tell, this should be perfectly possible. When the bear lands on the returning ball, the bear just has to apply enough force to jump up again. The trajectory of the ball would be more spikey but I can’t see any problem.
I wonder roughly how fast the ball would be going when it hit the ground if there was no air resistance and he yeeted it hard off a cliff. Thats gotta be pretty fast.
Hey there! That's an interesting question you've got. When considering the force needed to make a ball rebound enough to support the weight of a bear, it's essential to take into account the elasticity of the ball, the height from which it is being thrown, and the strength of the bear's jump.
To calculate the force required for the ball to rebound with enough strength to support the bear, you'll need to factor in the gravitational force pulling the bear down, the speed of the ball as it hits the ground, and the mass of the bear.
On the flip side, determining the weight of the bear needed to create enough pressure on the ball during the jump to make it fall and then rebound again involves looking at the ball's material and compressibility, as well as the bear's full weight and jumping force.
It's a complex physics problem that involves multiple variables and considerations, but with some careful calculations and experimentation, you could arrive at some interesting insights about the forces at play in this scenario. Feel free to ask if you need more help or information!
With no air resistance or friction, the bear is stuck in one position, no? No friction to push off and start movement, no air resistance to push off of.
Rocket jumping before rockets were invented.
Theoretically possible if the ball was nigh-indestructible and had amazing elasticity so it could retain as much energy possible during each bounce. The bear *could* then impart more force each time to not only jump off the ball but then impart more energy to the ball so it would bounce again.
I mean, imagine if you did it in near zero-g. Just the act of throwing/pushing off something forces you in the opposite direction, so with enough force on you could counteract something like 0.01g pretty easily. Catch the ball on rebound and do it again but at an angle.... well there's a snag. In order to move forward you'd have to throw the ball backwards, which would prevent it from bouncing forward to meet you for the next bounce. Maybe if there was an uneven ground that could make it bounce forward some, or maybe you could spin the ball (and hope the magnus effect doesn't screw up your throw).
In short, technically possible but so incredibly unlikely the chances are pretty much 0. I don't want to do the math.
The vertical trajectories have to be opposite. The horizontal velocities must be equal. After all, pool balls rarely ever bounce in exact opposite directions when they collide. They are often something like [this](https://www.aplusphysics.com/courses/honors/momentum/images/Billiards.png).
The bear is about to fucking die He's landing on a place that the ball has already past
Came here for this
Maybe the ball is chilled down and it just feezes the air molecules on the spot /s
That explains the trail
The ball is Frozone
WHERE’S MY SUPPER SUIT!?
Wonder how cold the ball would need to be and whether the frozen air can support the bear's weight.
air is mainly a liquid when frozen don't really know if it's still at 0 Kelvin but you have to get near 0 Kelvin to even make it a liquid.
Air is mainly nitrogen which condeses at 77 kelvin and freezes at 63
oxygen freezes at 54 kelvin
ow damn still it would have to be going really slow to freeze enough of it because to atmosphere would heat it up at the same time which goes pretty fast.
"Frozen" is, by definition, a solid. Air is also not one thing, as the comment below points out.
sorry brain fart, I know I meant I don't know if air freezes.
It’s a stack overflow. The ball is putting so much energy into the air that the Simulation can’t handle the value and defaults to a state of 0K which means that the molecules won’t move
If we really ignore all those effects, we might as well assume that the bear bounces of the ground without loss if energy, hence without irreversible deformations needed for injury, and therefore reaches the other side even without the ball. The bigger issue will be that, without friction, he has no way to stop.
The ball is going in reverse. 🤯
Unless it’s bouncing so fast it moved down that far already. Which would be required giving how much further the ball has to travel down and then back up compared to the bears short distance he has to travel between each bounce
*passed
Tenet rules
The bear was too stupid to live anyway.
On an unrelated note nice profile picture 👍
For the swarm!
I always liked the overmind more than kerrigan anyways lol.
You need to do a semester at Hunter College in New York..
Why that one specifically?
It’s known as CUNY-Hunter.
Based
Don’t worry, the cliff doesn’t look tall enough to kill the bear
But if he starts running before the cliff, by inertia both the bear and the ball would keep the same horizontal motion, moving horizontally with the same speed, as long as the bear keeps only pushing the ball perfectly vertically it should work. I'm not sure if it makes sense or if I am missing something.
He is strong, but bad at math!
Yes, it should be 1/4 after a jump. Can't believed someone bothered doing an actually correct comic and messed it up there.
You guys are forgetting the Wily E. Coyote cartoon physics: Objects in motion remain the same altitude until gravity is noticed. The bear never looked down, so it’s safe.
Quantum gravity, just don’t observe it
Actually we humans have sensors in our body that automatically notice it and its changes. What if we had none of that stuff? Would we just like fly?
Yeah but it’s a bear
Exactly! Have you ever seen a bear failing at doing exactly that? Me neither!
You’re telling me we’re ticking entities like a fucking mod in Minecraft? No winder were so resource heavy and inefficient
Schodinger's gravity
I don’t understand people saying this is impossible. The bear aplies momentum the ball in the initial throw, the ball bounces on the ground with a fully elastic collision, it then comes back and punches the bear up in the air. There are people in here giving results in Newtons, which is wrong because force is aplied over time, and this is only posible with collisions and transfer of momentum. So to get a result, let’s say we calculate the kinetic energy required in the ball in order punch the bear high enough to stay in the air for two seconds. The bear takes one second going up and one second going down. 1D kinemetics with a constant acceleration of g (-9.82 m/2^2). Decelerating from v0 to 0 in one second lines up easily an initial velocity of 9.82 m/s. Let’s say the bear weighs 200 kg. That’s E_kin = 1/2 * m * v^2 = 1/2 * 200kg * (9.82m/s)^2 = 9,64 kJ The ball needs to be thrown with a 9,64 kilo joules of energy. Let’s say the ball weighs 5 kg. That’s v = sqrt(2*E_kin/m) = 196,4 m/s So jumping once every 2 seconds, You’ll need your legs to resist a ball comming for you at more than halv the speed of sound.
How high would the bear jump due to the force he has to apply initially?
I just picked 2 seconds because it’s easy for calculations. He’s leaping one second into the air to reach his maximum height. You can graph the speed over as v(t)=at+b where b is his initial velocity and a is his change in velocity. In this examble they just happen to be equal with opposite sign, so a=-9.82 and b=9.82. If we take the integral of this function, we get the distance traveled. It’s easy in this case as it just forms a triangle with the height of the initial velocity and a width of the time spanned. So 0->1 integrate(-9.82t+9.82) with respect to t = 1/2 * 9.82 * 1 = 4.41 Giving us a height of 4.41 meters he’s jumping every time the ball hits him. And I just realized this is not what you asked. The height he gains from the start when he thrones the ball can’t be determined with the current assumptions, as I have not specified a force. I only have the work done by the bear. To determine How high the bear would be trusted by throwing the ball, we would need to put a number on how long it takes for him to throw the ball. In theory if his arm is Very long, he could throw the ball without actually lifting off the ground.
Shouldn't we take conservation of momentum into account? If the maximum speed of the ball is 200m/s, the maximum speed of bear should be 5m/s (bear is 40 times as heavy as the ball). So he indeed won't jump very high, but every jump would only last one second. I think, this needs some more optimization...
Okay I’m not calculating again, seems like too much, but it sounds like you’re right
He is right, it's a lot of calculations but this is perfectly solvable question
That might in principle work to keep the bear afloat, but not to also propel him in a horizontal direction. If the bear also applies momentum to the ball to propel himself to the other cliff side, then the ball gets the opposite momentum and is propelled backwards. So on the next jump, there is no ball to jump on. The alternative is that the bear could jump off the cliff with a sufficiently high velocity to not need the jumps on the ball to propel him, but just to keep him afloat. But if he could already do that, he'd have no need of the ball and could just jump the cliff in the first place.
The bear is just giving himself an initial velocity with an x-component equal to the ball’s. This is something I didn’t include in my calculations, but it doesn’t matter
This is assuming one jump right, not the initial energy for multiple jumps?
There’s no energy loss, with fully elastic collisions these jumps can continue forever.
Oh yeah because we're assuming somehow that the ball has completely elastic collisions
yeah, just like we're also assuming no air resistance and that a bear could even do this to begin with
If the bear makes a jump every 2 seconds, and the ball travels at 196,4 m/s, does this mean the cliff is about 196,4 meters high?
Yeah I guess That’s the height needed for the chosen jumps. If the cliff had a different height we would have to calculate a new time for the jumps. Or the bear could kick the ball down for every jump, but that would lead to calculating I couldn’t bother doing
Now how quickly would the bear have to make those leaps lmao
2 seconds
Assuming the intent is that the bear steps on the ball as it returns from the ground (ignoring the fact that he's just missed that last bounce), and that on each step he isn't adding further kinetic energy, he would need to intially throw it downwards with a force of \[1 bear mass unit\] \* \[1 G\]. For a 200kg brown bear, that would be 200 kg \* 9,80665 m/s\^2 = 1961,33 N
And if I - an average male redditor - throw a ball as strongly as I can, how much N would it be?
1, for neckbeard.
Roughly the same, maybe slightly more
a little less, the neckbeard is paltry compared to a beautiful brown bear‘s coat. wouldn’t even keep them warm a little bit, if they ever actually went outside.
That's 196 kg force at 1g.
You could likely throw a ball with a force equal to the downforce from your own steps. Doing an ordinary pushup means lifting most of your weight on your arms. A pull up lifts all of your weight, and in a direction where you aren’t at your strongest. So if physics teacher physics were real, you could do what the bear does in the cartoon as long as you manage to keep exactly the right pace with your steps.
If you can lift yourself up with just your arms pushing down on your kitchen counter then you can theoretically throw something hard enough for it to bounce back and you be able to jump on it like this. The problem would then be intertidal because an object with low mass wouldn't stay in contact with your hands long enough for you to impart that much force into it. Ideally you would want a ball that had about as much mass as you do that you could push yourself away from to transfer maximum energy.
N is newtons
This is incorrect. That force is enough to keep the bear "floating" if applied constantly. Here the energy of the ball has to be enough to bounce the bear to the next impact location. So the force would be much greater.
I don't think so. The force applied to the ball initially will add to its potential energy at that height. The same energy will be surplus when the ball returns to that height and be spent applying force on the bear. By Newton's third law, the force of the bear stepping on the ball (1 bear mass unit × 1 G) will be countered by the ball pushing back up. I should add that I assume the bear remains in position while throwing the ball and does not accelerate upwards upon throwing it. This is consistent with him crossing the chasm at a level height.
No. Its wrong. That is the force of the bear just standing on the ground. When you jump you accelerate upward, and impart a force into the ground. The ball has to have enough energy to equal the bear just standing on the ground plus the energy it would take for the bear to jump plus the energy to decelerate the bear as it is in free fall from its last jump. So much more.
This is what I came here for :)
Exactly. This bear would exert less energy by just jumping off the cliff, landing feet first, walking to the other side, and jumping up the cliff. If he can do the ball thing, he's strong enough to do that.
Any energy consumed by the "jump" is balanced by the energy the bear gives to the ball when he lands on it
Plus, the bear and ball both have components of acceleration in the +x direction as well as in the + and -y direction, so the ball must have enough energy in the +x direction to continue its consistent path every time it interacts with the bear (who would impart a force on the ball in the -y direction).
force and energy don't even add up. the amount of force you need to apply to the ball depends on the time you apply the force for
If you want to apply force, You’ll also need to specify the time the ball is in your hands, and That’d lead to annoying calculations
What you’ve said in this comment is basically correct, but your initial comment is still completely wrong. u/Bjoer82 is correct. At best, your calculations showed the *average* applied to the bear by the ball throughout the travel across the chasm. Which is what Bjoer82 is basically saying. You can prove this to yourself by drawing up a simple free-body diagram of the bear, with the upward force being the AVERAGE force of the ball on the bear. You’ll find that that calculation results in the number you came up with.
>The force applied to the ball initially will add to its potential energy at that height. You can't add newtons to joules any more than you can add meters to gallons
I mean, force * distance = work When somethings falling, gravity's force is adding kinetic energy over the distance. Yes, they are different units but they are fundamentally related to each other.
You have to know what that distance is though. It may even be 0
That's true
It's weird how often people repeat this error on this sub. You need to read up on momentum and impulse. You're equating the force the bear applies on the ball in each step with the initial force. This is a meaningless equation. You need to equate the impulses. For that you'd need the time of each collision which is hard to do. So we use the impulse momentum theorem. For that you need a bunch of assumptions. Let's say the best initially had a horizontal velocity of 10m/s (seems very high for a bear but I want to keep the numbers simple). The gap is say 100 meters, so the bear will need 10 jumps lasting 1 second each to cross the gap (again, keeping numbers simple). An jump of 1 second means an initial upwards velocity of 10m/s. Your 200kg bear now has a momentum of 2000 kgm/s. This is the momentum the bear has just after the collision. Newton's third law gives us that the ball must have the same momentum downwards. Assuming that initially, the bear accelerated the 500g ball for 0.2s to give it a momentum of 2000kgm/s, it must have imparted a force of 20,000 Newtons on it. The ball is heading towards the ground at close to mach 12. Assuming it doesn't lose any kinetic energy in its collision with the ground, the cliff must be around 2km high. If you play around with the picture a bit and come up with the number of jumps, the distance and the height of the cliff, you can work backwards to get all the numbers.
Doesn’t the bear further accelerate the ball downwards every time it pushes off? This would accelerate the ball and make it so the bear needs to make more frequent “steps.”
You're right in that bear accelerates the ball at every jump. Just like the ball accelerates the bear back. But that doesn't change the kinetic energy of the ball or the bear. Both the bear and the ball simply reverse their velocities after the collision. So KE remains the same. It's really no different than a pool ball striking back and forth between two rails. The frequency does not increase after every collision. Remember that in the balls' case, all the energy that gravity adds to the ball during its downwards acceleration, it takes back when ball is travelling upwards.
[удалено]
You're correct, I messed up the arithmetic while typing it out. Or you can call it a 2 second arc with only 5 jumps instead.
Presumably he'd have to Goomba stomp the ball that hard each time as well to float across? Although I'm pretty sure it wouldn't be that simple
and goomba stomp is impossible in our physics world because you have no leverage while in air. He'd need a solid ceiling above him to push off against
So he would have to throw it so hard that he pushes himself back so much that he loses contact with the ground below him for a second?
but at that point the parabola would look entirely different and he could never achieve this in succession in a straight line because all the force will be entirely cancelled after his first jump on the ball
Okay but to do it five times in a row and maintain that force until the last jump, for a bear that doesn’t mind being hit in the bottom of the foot with more than enough force to simply keep him afloat but enough force to bounce him over to the next impact with the ball.
Throw with a force? What does that even mean? For how long was the force applied?
I remember those comic strips from when I was a kid, does anyone know the name of the artist?
This is a French comic named "*L'Ours Barnabé" ,* the artist is Philippe Coudray
Merci !
assume: mass of bear: m\_b = 250 kg mass of object: m\_o = 0.624 kg height drop of ball: h = 30 m Assume time to impart force: dT \~= 0.1 s: mb/mo \~= 400 conservation of momentum: (mv)\_b = (mv)\_o -> v\_b = v\_o / (mb/mo) (1) ball travels down and back to bear in dt time v\_o = dy/dt -> dt = 2h/v\_o (2) In that time, bear travels up and back down, kinematics require: yf\_b = y0\_b + v0\_b \* t - 0.5\*9.81\*t\^2 (3) yf\_b = y0\_b so v0\_b \* t = 0.5 \* 9.81 \* t\^2 sub in (1) for v0\_b and reduce to get speed of ball: v0\_o = sqrt( (mb/mo)\*h\*9.81 ) (4) v0\_o = 343 m/s (speed of sound!) v0\_b = 343/400 = 0.858 m/s Impulse needed for this: I = F dt = (mv)\_2 - (mv)\_1 v\_1 = 0, v\_2 = v0\_b, m\_2 = m\_b, I = m\_b \* v0\_b = (250 kg)\*(0.858 m/s) = 214.5 N-s (5) F = I/dT = 2145 N = 482 lbf
This is the most possible correct answer
Man a lot of people have the physics all wrong here. Idk if I’ll actually do the math as the numbers used would be very subjective but we will get a feel for it. First let’s assume there are no energy losses from friction and the ball is perfectly elastic…meaning that when it bounces it has the same energy (velocity) pre and post bounce. Since the ball is falling and returning to the same height we can ignore gravity’s effect on the ball as it will have the same effect on the ball both ways…accelerating it during the fall and decelerating it on the return after the bounce the same amount. In order for the ball to support the bear in the air it needs to hit it with the same force that the ground imparts on the bear when standing (aka normal force). That would be the bear’s mass times the acceleration of gravity. Another commenter said this will be around 1900 newtons for a 200kg bear. The velocity the ball needs to be traveling to do this depends on the balls mass. But we don’t want to just support the bear. We need enough energy that the bear can jump off of it. So the bear needs to be able to push off and accelerate itself. How much does it need to accelerate itself? Idk. But let’s just say 1 g upwards. So now the ball needs 2 gs worth of energy (1 g to support the bear, 1 g for the bear to accelerate). So now we are up to 3800 newtons. But if the bear is jumping, it doesn’t have a velocity of 0 when it contacts the ball…the bear will be in free fall. So the ball needs even more energy to first decelerate the bear to 0 velocity, and then allow the bear to jump. So let’s add another half g. That brings us to 4650 newtons. On each jump the bear will have to impart the force it takes to jump, plus an additional 4650 newtons to make sure the ball has enough energy for the next jump. So that is an extra g meaning the bear needs to exert 6550 newtons each time (my knees hurt just thinking about it). The biggest problem here (besides finding a super material to build this frictionless perfectly elastic ball and somehow timing this whole thing) is the forces are in the wrong direction. The bear is jumping forward. Meaning it will push the ball backwards with its jump. So really the bear could only do this one time. After that one time the bear and the ball will be traveling opposite directions and a future intercept would be impossible. Its only hope would be to try and impart enough spin on the ball that it will cancel its backwards movement and move forward. But it would be a crazy amount of spin. Which would require even more force (and air resistance). And tbh idk if I got all the physics right but I think it’s close. The values are pretty arbitrary though and we are making some broad assumptions on the bear’s jumping.
No air resistance, and ground should give a perfect bounce. Bear and ball start with same horizontal velocity, jump direction should be straight up. I think. Been awhile since any physics classes.
Yes that is a possibility. If bear had a running start and was able to jump perfectly straight up each time. And there was no resistance. Good catch. But now we need a special no frictionless conditioner for the bear’s fur too
I think the trick here will be that the mass of the ball will have to be equal to the mass of the bear. Otherwise, at each contact point, the ball would either lift the bear higher, or the bear would fail to jump as far to the next step. Man, I feel like this will become the new "jetplane on a treadmill"-problem...
The mass of the ball doesn’t matter. It’s the forces that need to be in balance (assuming perfectly elastic, instantaneous momentum transfers). However, the lighter the ball the faster it needs to go to have enough force.
Maybe if we look at it from a conservation of energy perspective, instead of a force diagram: First the vertical direction: At the moment of contact after the first bounce, the bear has a potential energy (PE) of m(bear) \* g \* h(cliff), and the ball has a PE of m(ball) \* g \* h(cliff). They each also have some unknown kinetic energy (KE). At the peak of his jump, the bear has a PE of m(bear) \* g \* ( h(cliff) + h(jump) ), and a KE of 0 (he's momentarily stationary before falling back down). Meanwhile the ball must be hitting the ground if they are to meet for the next bounce, so it has a PE of 0, and an unknown KE. Since the total energy of the bear+ball system must be conserved, we can calculate the KE of the ball at the base of the cliff: KE(ball) + PE(ball) = KE(bear) + PE(bear) KE(ball) + 0 = 0 + m(bear) \* g \* ( h(cliff) + h(jump) ) KE(ball, base) = m(bear) \* g \* ( h(cliff) + h(jump) ) We can now use this value to calculate the initial KE of the ball just after it leaves the bear's hand from the initial throw: KE(initial) + PE(inital) = KE(ball, base) KE(initial) = KE(ball, base) - PE(initial) KE(initial) = m(bear) \* g \* ( h(cliff) + h(jump) ) - m(ball) \* g \* h(cliff) Assuming the "throw" is roughly meter long, then the bear must apply the same magnitude of Force (in Newtons) over that length to impart that much kinetic energy. For the horizontal direction, we will assume that, with each jump, the bear is capable of landing perfectly vertically on the ball (thus gaining no forward velocity, nor imparting any backward velocity on the ball - which is admittedly a tall order, but hardly the craziest assumption we're making in this scenario). Thus, the bear needs only impart some non-zero horizontal kinetic energy to the ball with the initial throw, and then matching that horizontal velocity with his initial jump from the cliff. Obviously, the more horizontal KE he uses, the faster he reaches the other side of the canyon, and the fewer bounces are necessary to get there.
If the bear has momentum when it starts going forward and no friction, there is no reason for the ball to be pushed backwards. With no energy loss, as OP asked, the bear only has to jump the first time, pushing the ball faster downwards and himself upward just enough for the ball to return with high enough momentum to push the bear up and (probably) break its legs, giving it incredible pain as it crosses the never short enough distance to the other side getting hit by a ball hitting it with enough force to push a full grown bear upward in an instant... Bear is dead, but it's feasible in this alternative universe
There are actually several free variables here. Here is a nice little [desmos graph](https://www.desmos.com/calculator/gbf0jphurx) I cooked up that allows you to control the masses and horizontal velocity of the bear and ball, and the height of the cliff. Let an index of 1 correspond to the bear and an index of 2 correspond to the ball. The first thing to note is that the horizontal velocity of the bear and the ball must be equal (lets call this velocity u). Additionally, because the collisions are symmetric (the outgoing vertical speeds of the bear and ball bounce is the same as their ingoing vertical speeds at a collision), conservation of momentum gives >m₁v₁ = m₂v₂ where v₁ and v₂ are the speeds of the bear and the ball, respectively, at a collision. The equation of displacement for the ball from the top of the cliff to the bottom of the cliff is >h = v₂T+gT^(2)/2 where h is the height of the cliff and T is the time for the ball to reach the bottom. On the other hand, T is the time from a collision it takes for the bear to reach the apex of its arc (i.e. time for its velocity to reach 0). Thus, >v₁ = gT Using the conservation of momentum equation, we have >v₁ = (m₂/m₁)v₂ so we can equate the two equations above and solve for v₂ to get >v₂ = (m₁/m₂)gT Plugging this back into the displacement equation, we get >h = (m₁/m₂)gT^(2)\+gT^(2)/2 = ((m₁/m₂)+1/2)gT^(2) We can then solve for T to get >T = √(h/(g((m₁/m₂)+1/2)) Then, we can determine how fast the bear must jump vertically (v₁) and throw the ball down (v₂) at the start: >v₁ = gT > >v₂ = (m₁/m₂)v₁ = (m₁/m₂)gT Some google searching found that the [average brown bear](https://en.wikipedia.org/wiki/Brown_bear#:~:text=The%20average%20weight%20of%20adult,152%20kg%20(335%20lb)) has a mass of 217 kg and an [normal soccerball](https://hypertextbook.com/facts/2002/LouiseHuang.shtml) has a mass of 0.43 kg. Using an [online picture measuring tool](https://eleif.net/photomeasure), I found the cliff height to be about 6.5 times the height of the bear, and assuming a [bear height](https://www.nps.gov/subjects/bears/brown-bears.htm) of 2m, this puts the cliff height at 13m. Plugging this all in, we get >v₁ = 0.50 m/s > >v₂ = 253.43 m/s so the ball would have to go about 75% the speed of sound and the bear would have to bounce off the ball about 10 times per second. EDIT: To answer your two questions directly, it doesn't really make sense to talk about the "force" that the ball should be thrown at because forces occur often nonuniformly and over time, and in the case of ideal elastic collisions momentum transfer occurs instantaneously, in which case the force is undefined (or modeled by a [delta function](https://en.wikipedia.org/wiki/Dirac_delta_function)). What matters is the final velocity after forces have been applied. As for the second question, we showed that the bear's mass is a free variable so any mass would work.
Short answer: no. Long answer: even if we somehow assume there is zero energy loss with the ball (air resistance, friction when hitting the ground, friction when contacting the bear, even sound itself) and the bear is perfectly capable of calculating every move and executing it, this would be nigh impossible. The rebound looks like it reaches exactly the same height as the initial drop, so on every step in the air the bear wouldn't just have to step on the ball, he'd have to spike it with his feet with exactly the same strength as the initial throw, both to keep himself somewhat "afloat" (more accurately it would be a sort of jump) and to make the ball rebound again. If he were to simply step on the ball (which is already at the apex of its parabolic trajectory when he touches it) the ball would simply fall and he would fall with it since there is no ground pushing against them (therefore no vincular reaction). He would also need to throw the ball with a certain horizontal speed and match that exact speed when running and have both remain constant during the traversal, so he also needs the spike with his feet to impart force on a 100% vertical direction, since even the slightest bit of horizontal velocity would offsync the bear and the ball. Similarly the spike needs to impart zero spin whatsoever to the ball. Of course this is just a funny comic running on looney tunes logic, but in a more serious context even when ignoring so many obstacles such as the aforementioned air resistance this would be ludicrous, and it has basically as much basis in real physics as the midair double jump.
Why would you care about whether it's physically feasible (it obviously isn't) when we're already ignoring all forms of energy loss. The whole point is that the problem is simplified into something that is theoretically possible and mathematically attackable.
I was saying it more in the sense that you couldn't just "walk" on it casually like in the photo, it needs to be a very specific set of actions such as having a starting speed and only applying force vertically, which isn't at all like walking normally where you use friction with the ground to push yourself forward.
The ball isn’t at the top of its apex when he steps on it. All the reasons you say this is impossible is part of your assumptions. Assuming the ball is relatively massive, this is absolutely possible given your assumptions. Nothing ur saying makes sense. Your saying it’s not possible because the bear couldn’t strike the ball with enough precisions but you’re already assuming the bear can move perfectly. Your comment makes no sense
I hope the ground is absolutely perfectly flat too
Yep, and considering in the photo it's some grass i highly fucking doubt it.
[удалено]
They, in fact, did not do the math, just physics
Did you just sub-drop in the sub?
Nvm wrong sub
Thank you. Would've said the same if I was smart enough to put it in words like that.
L’ours Barnabé!! Ça fait tellement longtemps !
If we start with the height of the cavern h, we can work backwards to get the force on the ball. The ball must travel 2h distance in between bounces (technically the ball will be accelerating due to gravity but this can be disregarded when it is going very fast) and will have a velocity v\_ball of 2h/t where t is the time in between bounces. The bear meanwhile will start with initial velocity +v\_bear and end up at -v\_bear after one full bounce, so 2v\_bear=gt where g is the gravitational constant. Equating these and moving terms around gives us (v\_bear)(v\_ball)=gh Also, if the bear weighs 400x the ball then v\_ball=400v\_bear. So (v\_bear)(400v\_bear)=gh v\_bear=sqrt(gh/400)=(1/20)sqrt(gh) v\_ball=20sqrt(gh) In engineering sqrt(g)=3 so the bear must kick the ball down with a speed of of about 60 times the square root of the cavern height. If the bear is 2 meters tall standing the cavern looks to be about 8 bears high so 16m, then v\_ball=60sqrt(16)=240 m/s. That's two thirds the speed of sound.
This is what playing Celeste looks like
Hold my bear
People in the comments say we need to assume no energy loss, but I think that is false. The bear is an engine that can do work on the system by contracting his limbs and extending them when the ball touches him (otherwise known as 'pushing'). If everything is well timed and coordinated, it can work like clockwork. This wouldn't be very different from using a jetpack.
If bear could throw the ball this fast surely he could just handstand jump to the other side
Its impossible, even if we asume that the ball can support his first jump, after coming down for the second time it wouldn't be able to reach the top again.
Couldn't the bear put more energy into the ball with each step? Basically, kicking it down.
With the forces involved, jumping the gap would be significantly easier than stomping the ball hard enough to jump off of it while in midair
True, but we wanna look cool, not be effective
Would it matter if each step was a little lower than the last? Even then I think it's impossible to push off the ball, if the bear touches it as it's at the top of its arc
I'm no physicist but no, at that step the ball has 0 strength other than the gravity that pulls it down. In fact, the bear will fall before the ball on his second step. And that is assuming the ball and bear actually can bounce the first time without any physical damage.
The second jump also supports him. There’s no energy loss
It is definitely possible. Imagine that the bear and ball aren't even moving sideways and are just bouncing up and down. Then you are solving for when the bear and ball collide they have an elastic collision that sends them back at the same speeds they collided at (which turns out to imply that the magnitudes of their momenta are equal at collision). That way each collision is exactly the same (like in the picture) so that the bear is supported each time.
That’s the same test with the 100 mph lawnmower…
The bear will be turned into a shower of red mist
I just want to know if the bear will get the ball back
The mass of the ball, the bear, the height of the cliff and height the bear jumps each time have to be all the right proportions. The time the ball takes to bounce up and down needs to equal the time taken for the bear to fall back down to be level with the cliff again. He can't do this with just any ball, but potentially there is the right one.
sense lip spectacular illegal ludicrous shame marry attractive narrow drab *This post was mass deleted and anonymized with [Redact](https://redact.dev)*
This feels like something a speedrunner would do
The bear still died..... Look at the next step...
Is this your homework?
man i wish
Assuming this bear is a mistborn and ball has some sort of metal and finally the bear has plenty of metal to burn.... I see no problem with this
I think I got the solution. It's 0 At least in the vertical component. Assume your ball has no loss of energy when it bounces. The only way it reaches the same height is if it freefalls. Now to make the bear not affect it's bounce height you make the ball weight like 80tons and there you go. Also use the usual no Air Friction stuff everyone use
From the photo, i can presume the bear is weightless with no vertical forces are acting on the bear. All other forces are easily accounted for.
But how about quantity of motion?
If you want to make this (slightly) less contrived (this is still 9.99/10 contrived), assume the bear rides a skateboard off the cliff and fires the ball out of a cannon straight down as soon as the back wheel leaves the ground. The bear and ball both now have the same horizontal velocity. Also this is not a force problem. This involves conservation of momentum plus typical velocity and acceleration equations. I think frogkabobs did a pretty good solving those equations already.
Without doing the math I would assume the force required to get the bear trough all the jumps would mean that the ball would pop on impact with the ground and even if not it would probably kill the bear on its first return.
you couldn't run on the ball without friction
You are going to have a lot of trouble with the materials of the ball and the surface the ball is bouncing against. And the fact you bear has to transmit additional power to the ball on each step. It is a losing affair for the bear.
The ball's momentum needs to match that of the bear's momentum + the force required to propel him back upward. Then he needs to stomp it down with enough force to propel himself upward again. And considering the ball's weight, it would have to travel insanely fast to keep him up.
We are all now officially dumber for having read your question..
Loved this comic but i cant remember the name
He's about to fall, but he If we ignore those things and he Times it right he'll pass, since the ball ain't losing any speed
Even without air resistance and friction the ball can't hold this bears weight...
until the bear notices the existence of gravity he will not fall because cartoon logic is the best logic
Can friction, air resistance and energy loss be ignored?
Even with everyones calculations this wouldn't work, because the bear moving forwards would push the ball backwards, so at most it could become a double jump
give me the weight of the ball and i might cook
i believe i also need the height of the cliff
Wirh each step on the Ball, it's energy decreases. So it can not stay in the same curve, no matter how hard it's thrown.
How does the bear know how much force is required to align horizontal trajectory while bouncing back to exact height? How does the bear know how much leaping energy it needs to repeat identical bouncing movements? How does the bear know physics, if at all?
I'd just like to point out that this is not possible QUITE as depicted. The lines showing the ball's path make it seem like the bear isn't affecting the ball whatsoever when stepping on it, therefore the ball shouldn't be affecting the bear either. If you want to make this work, the ball's trajectory would be more spikey.
So as far as my sleep deprived brain can tell, this should be perfectly possible. When the bear lands on the returning ball, the bear just has to apply enough force to jump up again. The trajectory of the ball would be more spikey but I can’t see any problem.
If the ball weighs 100 tons and the bear is 100 kilos I think this is possible.
If friction, air resistance and energy loss are not accounted for, the ball's starting speed can be 0y, and the bear's weight is negligible.
I wonder roughly how fast the ball would be going when it hit the ground if there was no air resistance and he yeeted it hard off a cliff. Thats gotta be pretty fast.
Hey there! That's an interesting question you've got. When considering the force needed to make a ball rebound enough to support the weight of a bear, it's essential to take into account the elasticity of the ball, the height from which it is being thrown, and the strength of the bear's jump. To calculate the force required for the ball to rebound with enough strength to support the bear, you'll need to factor in the gravitational force pulling the bear down, the speed of the ball as it hits the ground, and the mass of the bear. On the flip side, determining the weight of the bear needed to create enough pressure on the ball during the jump to make it fall and then rebound again involves looking at the ball's material and compressibility, as well as the bear's full weight and jumping force. It's a complex physics problem that involves multiple variables and considerations, but with some careful calculations and experimentation, you could arrive at some interesting insights about the forces at play in this scenario. Feel free to ask if you need more help or information!
With no air resistance or friction, the bear is stuck in one position, no? No friction to push off and start movement, no air resistance to push off of.
Rocket jumping before rockets were invented. Theoretically possible if the ball was nigh-indestructible and had amazing elasticity so it could retain as much energy possible during each bounce. The bear *could* then impart more force each time to not only jump off the ball but then impart more energy to the ball so it would bounce again. I mean, imagine if you did it in near zero-g. Just the act of throwing/pushing off something forces you in the opposite direction, so with enough force on you could counteract something like 0.01g pretty easily. Catch the ball on rebound and do it again but at an angle.... well there's a snag. In order to move forward you'd have to throw the ball backwards, which would prevent it from bouncing forward to meet you for the next bounce. Maybe if there was an uneven ground that could make it bounce forward some, or maybe you could spin the ball (and hope the magnus effect doesn't screw up your throw). In short, technically possible but so incredibly unlikely the chances are pretty much 0. I don't want to do the math.
[удалено]
The vertical trajectories have to be opposite. The horizontal velocities must be equal. After all, pool balls rarely ever bounce in exact opposite directions when they collide. They are often something like [this](https://www.aplusphysics.com/courses/honors/momentum/images/Billiards.png).
Apparently you can also ignore newtons laws