A ball is dropped from 44" and bounces to a height of 19". How long does it take to come to rest?

First we need a consistent frame of reference; I’m going to use the bottom of the ball since this is where it contacts the floor. So the original height is 42” from the bottom of the ball and the rebound height is 15”.

Potential energy is mgh, meaning that it’s directly proportional to the height. This means the percentage of height lost is also fixed. With each bounce it’s losing about 64% of the height of the subsequent bounce.

64% of anything nonzero will never actually reach 0. But after 4 bounces, the ball will bounce less than an inch off the ground, which is an okay approximation for resting.

Now how long does all of this take? The relevant kinematics equation is d = v0*t + .5*a*t^2. a is 386 in/s/s on Earth. v0 is initial velocity and is 0 for each leg of the trip; velocity is 0 at the exact moment that a change in direction occurs. d changes for each leg of the trip.

First the initial fall of 44”: 44” = 0*t + .5 * 386 * t^2; t = .47 seconds
Then the upward bounce of 15”; just change d to 15. t = .28 seconds
Back down the 15” to the ground is the same amount of time as it took to rise 15”. .28 seconds.
The next rebound is 36% of 15”, or 5.4”. t = .17 seconds
Back down: t = .17 seconds
Next rebound is 1.9”; t = .10 seconds
Back down: t = .10 seconds
Next rebound is .68”; t = .06 seconds
Back down: t = .06 seconds.

And like I said before, this is where I stop because t is getting to be negligible. Add together all the t’s: 1.69 seconds.

Lots of rounding, lots of corners cut. Not sure if that’s what you were looking for.

3.61 secs.

@ucme What’s the basis for that answer?

@Mariah That seems like a realistic number. One observation that I have is that 0.28, 0.17, 0.10, and 0.06 form the start of an infinite series that approaches zero. Does this series converge, as to place a hard limit on the amount of time?

@PhiNotPi I’m glad you ask young man, it’s really very simple, I emulated the action described in your question with a stopwatch in hand.
Of course, it depends on the density of the floor, in this particular instance wooden flooring was used.

@ucme How were you able to make sure that the bounce reached the correct height?

We’d want to be more precise if we were going to analyze this with an infinite series. The series is actually defined as such:

sqrt(44/193) + Summation(2*sqrt(44 * .36^i / 193)) from i = 1 to i = infinity

I don’t actually know the rigorous way to find what such a series converges upon! But what I usually do is just run a large number of iterations through my calculator until the number stops changing by any significant amount. When I do this I get about 1.91. Sorry, I know that’s really unsatisfying. :(

@Mariah After a certain point, the series becomes a bad model of the situation. If it followed the infinite series exactly, the rate of bouncing would increase continuously, until it reaches a speed of infinity bounces per second about 1.89 to 1.91 seconds into the experiment.

Yeah the number of bounces approaches infinity as the height of the bounces approaches 0. It actually is a good model of your situation as described; a fixed loss of energy per bounce. The trouble is that that situation is not a good model of reality, once we get into the nitty gritty details.

@PhiNotPi I confess a rough estimate was used, but I can repeat the process utilising a tape measure if you like, I really don’t mind.

@Mariah, well, then what is a good model of the actual situation?

@ucme I still don’t know if I understand. I know that you can find a 4” diameter ball and drop it from the correct height, and measure the time. But what I don’t understand is how you can replicate the 19” rebound, since that is influenced by the “bounciness” of the ball, and I doubt that we are using the same type of ball.

My guess would be that, once the bounces got very low, the ball would not be deformed as much by the impacts, and the bounce height would drop off more rapidly. The ball would actually come to rest instead of vibrating very quickly.

If we want to get really involved, we could say the Planck length is the smallest possible bounce and consider the ball at rest when the bounce gets to that height. But at this point the bounces would be taking a ridiculously small amount of time and we would still get an answer very very close to 1.91. This is why we don’t usually take quantum or relativistic effects into account when we do kinematics problems – Newton’s laws are a good enough approximation!

One part of the problem is the force of G earth-gravity as a constant downward accelerating force. The perfectly elastic ball would approach the asymptotically flat bounces as described above.

Another is inefficiencies in the elactic properties of the ball. Any reduction in height beyond that explained by G is from friction in the bounce, a damping factor on the bounce/spring constant of the ball itself.
This aspect is why it may be difficult for @ucme to collect empirical evidence that actually matches the second bounce height. On the other hand if it does match I’d consider his observed practical resting time a substantial data point.

@dabbler IIRC gravity itself does not reduce the bounce heights. If we assume perfectly elastic collisions and no friction, the ball would bounce back up to the same height each time. No loss of energy = same height attained. The inefficiencies in the ball are built into the problem by assuming that this does not occur. However, they’re not modeled in a perfectly realistic way, which is why we’re getting a ball vibrating infinitely quickly at the end.

Guys @ucme is most definitely pulling our legs.

Actually, one thing that we should consider, rather than the Planck length, is the speed of sound in the material. Once the ball hits the ground, there is a certain amount of time required for the top of the ball to “know” that it has hit the ground. If bounces are happening faster than this amount of time, the ball itself it simply vibrating, and the bouncing motion becomes sound waves within the material.

@PhiNotPi Of course it depends on the bouncebackability (what a fantastic word) of the ball, as does the density of the floor, carpeted/vinyl/wooden.
I used a rubber ball, quite heavy & so yeah, the first bounce would be difficult to replicate.
@Mariah My goodness gracious me no, this being in general & asked by a mod…I shudder to think.

by the way, I am not directly responsible for any moderation on this thread

@ucme Okay, that clarifies things.

@Mariah I think that we need to account for the curvature of spacetime. I mean, why not?

Well, anyways I guess this problem is about as solved as it could be. Now for the next level! If X is the drop height, and Y is the height of the second bounce, what is the formula for f(X,Y) for the length of time?

sqrt(x/193) + Summation(2*sqrt(x* (y/x)^i / 193)) from i = 1 to i = infinity

If x and y are given in inches.

sqrt(x/193) + Summation( 2 * sqrt( (x * y) / (x ^ i) / 193) )

Is this correct?
Then:

sqrt(x/193) + 2* Summation( sqrt( (x * y) / (x ^ i) / 193) )
sqrt(x/193) + 2* Summation( sqrt( (x * y) / (x ^ i) ) / sqrt(193) )
sqrt(x/193) + 2/sqrt(193) * Summation( sqrt( (x * y) / (x ^ i) ) )
sqrt(x/193) + 2/sqrt(193) * Summation( sqrt( y ) * sqrt((x) / (x ^ i) ) )
sqrt(x/193) + 2*sqrt(y)/sqrt(193) * Summation( sqrt((x) / (x ^ i) ) )
sqrt(x/193) + 2*sqrt(y)/sqrt(193) * Summation( sqrt(1 / (x ^ i-1) ) )
sqrt(x/193) + 2*sqrt(y)/sqrt(193) * Summation( 1 / sqrt(x ^ i-1) ) )
sqrt(x)/sqrt(193) + 2*sqrt(y)/sqrt(193) * Summation( 1 / sqrt(x ^ i-1) ) )
(sqrt(x) + 2*sqrt(y))/sqrt(193) * Summation( 1 / sqrt(x ^ i-1) ) )

Is this correct, can you double check all of this? If I remember correctly, the summation at the end has no closed form.

Right at the start you pulled y out of an exponent that it’s a part of. Don’t do that!

@dabbler Wise words indeed :)

@Mariah Did you ever edit your answer? The very first line of the answer was simply me adding in parentheses. I didn’t pull anything out of the exponent.

Yeah my bad, I threw parenthesis around (y/x) because the exponent is ambiguous without them. Sadly I think you started your proof using my erroneous initial answer.

Take 2:

sqrt(x/193) + Summation(2*sqrt(x* (y/x)^i / 193))
sqrt(x/193) + 2* Summation(sqrt(x* (y/x)^i / 193))
sqrt(x/193) + 2* Summation(sqrt(x* (y/x)^i) / sqrt(193))
sqrt(x/193) + 2/sqrt(193) * Summation(sqrt(x* (y/x)^i))
sqrt(x/193) + 2/sqrt(193) * Summation(sqrt(x)* sqrt((y/x)^i))
sqrt(x)/sqrt(193) + 2 * sqrt(x) / sqrt(193) * Summation(sqrt((y/x)^i))
sqrt(x)/sqrt(193) * [1 + 2 * Summation(sqrt((y/x)^i))]
sqrt(x/193) * [1 + 2 * Summation( sqrt(y^i / x^i) )]

Yeah all of that looks good to me. Now to remember ways to prove convergence or divergence…

Well the series definitely diverges if y >= x because of the rule that the sequence must approach 0. But in the context of the physics problem, that won’t happen unless there is no friction whatsoever, and that makes sense because the ball would just keep bouncing forever.

However, the fact that it does approach 0 when y<x does not prove that the series converges. I forget my calculus III class where I learned to do these proofs. Lemme refresh myself real quick…

Oh yeah! This is a geometric series. Each term is a ratio of the previous (from your original example, each term is .36 times the size of the previous). And geometric series converge. I even know how to find what they converge upon! Gimme another second.

It’s not a geometric series because of the location of the square root!

Oh damn. I forgot about the square route.

I gotta go to dinner. I’ll rejoin you after!

Okay. We can prove convergence by a comparison test.

This is a geometric series put beneath a square root. The square root makes it smaller than a geometric series (for large numbers). A geometric series converges, so this smaller function must also converge.

Unfortunately I still don’t know the proper way to prove the value it converges to. I’ve never taken a class that covers that! I think the one I’m taking in the spring of next year might though. :) I hope so, because I love this topic.

@Mariah Since the number inside of the square root is always less than one, the square root actually makes each term bigger.

sqrt(5^2 / 7^2) > 5^2 / 7^2

Damnit, hasty again. You’re right. Let me just think out loud for a moment…

Limit comparison test. If we have two functions a(n) and b(n) lim n->infinity a(n)/b(n) = c with 0<c<infinity, then both a and b converge or a and b diverge.

Let’s define a(n) to be what we have and b(n) to be the geometric series that is the same thing minus the square root sign.

sqrt((y/x)^n)/(y/x)^n
(y/x)^(n/2)/(y/x)^n
(y/x)^(-n/2)

lim as n->infinity = 0. Damn.

It’s not a p series. It’s not a geometric series. It’s not an alternating series. Umm…I can’t think of any other methods I learned to prove convergence! But I took this years ago…

Summation(sqrt((y/x)^i))
substitute y/x for z, to simplify.
sqrt(z^i)
z^(i/2)

This is similar to a geometric series. Instead of z^1, z^2, z^3, it’s z^0.5, z^1, z^1.5.

This can be broken up into two series: 0.5, 1, 1.5… = 1, 2, 3, combined with 0.5, 1.5, 2.5…
Summation(sqrt((z)^i)) = Summation((z)^i / sqrt(z)) + Summation((z)^i)
Summation((z)^i)/sqrt(z) + Summation((z)^i)
Summation((z)^i) * (1 + 1/sqrt(z))

z/(1-z) * (1 + 1/sqrt(z))

!!!!

Is this right?

Hmm no not quite. you pulled the sqrt(z) out of the summation in the second to last step. You’re on an awesome, creative track though. Gimme a minute.

EDIT: Wait no sorry. That’s a constant. You’re good.

sqrt(x/193) * [1 + 2 * (y/x/(1-y/x) * (1 + 1/sqrt(y/x)))]

now to simplify

@Mariah The sqrt(z) is a constant. Can’t that be pulled out of the summation?

Yeah see my not-quite-quick enough edit. I still think there’s an error somewhere though because when I repeat with the example of z = .36 I don’t get the answer I got before.

Note to all: edits are not viewable to the other jellies until those jellies refresh the page.

I hope to god that there wasn’t an error in the original summation.

Wait a minute. Hot damn! You’re right! I forgot the summation formula you gave above wasn’t the full thing. Plug .36 in for y/x in your comment above and you totally get 1.91. Fucking awesome work dude.

As for the original summation, let me double check that now.

Okay so let me just talk out my original summation.

The first thing to know is that 193 comes from .5*386, 386 being g in in/s/s. The first term is sqrt(x/193) because this is the case of i=0. I didn’t just include it in the summation, because this is the only term that happens only once – the ball falls 44”, but it never bounces back up 44” like every other bounce height.

So having covered i=0, we start our summation at i=1.

The kinematics formula that this all came from was d = v0*t + .5*a*t^2. But we can simplify this to d = .5*a*t^2 because v0 will always be 0. .5*a is 193. So t is always sqrt(d/193). d needs to be multiplied by a number 0<z<1 in each iteration. This can be accomplished by raising that number z to i in the summation. And that number z is the percentage of the previous height that the current height is. Since this is to remain constant for all bounces, it can be found as the ratio of the first bounce to the original drop height of the ball, given in your problem statement as y/x.

For each height, the ball needs to both rise up to that height on the bounce, and also fall back down the ground. Early physics teaches us that this rise and fall happen in the same amount of time. So multiply each term by 2.

And that’s how we end up with Summation(2 * sqrt(x*(y/x)^i / 193). Agreed? It checks with the numbers I found by hand originally (with some rounding error).

@Mariah That seems good, and it is backed by experiment. When I performed the original trial, I’d reckon to say that it took a little more than 1.5 seconds to stop bouncing, although that was not an ideal bouncy ball. If there were an error in the formula, the answer would have been orders of magnitude away from real life.

[High five] we dunnit again my friend.

Hey, you. Yeah, you. If you don’t want to follow the math in this thread, just go ahead and slap a GA here. I can vouch for its brilliance.

Thanks a lot @Mariah for all of your help.

Please continue coming up with these questions, they are the highlight of some of my days…no I am not a boring person!

Useful links:
Link1
Link2

@prasad, Thanks for the links. The infinite number of bounces in finite time reminds me of the problem related to Zeno’s paradox of the turtle racing against Achilles.

My guess is that the geometric series model may work for the first few bounces but that it will become less accurate for small heights. If you take a small rubber ball and very gently squeeze it, nothing happens. There seems to be a threshold that needs to be reached before the ball will deform. I would think that once the height gets small enough, the ball will not bounce any more.

@prasad Good stuff, well found.