4 ms·
Where's the spoiler?
by pero 16y ago
Where's the spoiler?
- btilly 16y agoHere is a spoiler. Energy = force * distance After each bounce you are left with 60% of its energy, so it comes back up 0.6 times as high as the previous bounce. Distance falling in time t is proportional to the square the time, so each bounce takes sqrt(0.6) times as long as the previous bounce did. Thus the timing of the bounces forms a geometric series. It is well known that the sum of such a geometric series is 1/(1-r). In this case r = sqrt(0.6) which is roughly 0.774596669241483 and so from the time it first hits the ground to the time it it finishes bouncing is approximately 4.43649167310371 times as long as the time for the first full bounce. But we didn't start with a full bounce, we dropped the ball. Thus we start with a half-bounce, followed by a full bounce that takes 2 * sqrt(0.6) times as long, followed by the rest of the sequence. This works out to be 7.87298334620742 times the time it took to initially fall to the ground the first time. Hopefully I haven't made any silly mistakes. If I have, correct the error and the general analysis is correct.
- greenlblue 16y agoSame mistake as everyone else. Geometric series means infinitely many bounces and infinitely many bounces means it never stops bouncing. Everyone is making the same logical fallacy.
- btilly 16y agoI refer you to Zeno's paradox for an example of how a geometric series can allow an infinite number of things to happen in a finite time. In this case the time taken forms a geometric series, and the total time taken is the sum of that geometric series. Which means that, for the same mathematical reasons that let Achilles catch the tortoise, it stops in finite time.
- greenlblue 16y agoI am familiar with Zeno's paradox and all other things Zeno and infinite series summing to finite things but there is fallacy here that nobody seems to get. Yes the time taken is indeed a geometric series but if the time taken is a geometric series, an infinite one at that, then that means there are infinitely many bounces, no? So if there are infinitely many bounces how can you claim the ball stops bouncing? You are confounding two things, air time which is indeed finite because it forms a geometric series, and the number of bounces. You can not have an infinite geometric series to calculate the time and only have finitely many bounces.
- sswam 16y agoThe ball bounces infinitely many times, in a finite period of time, and then stops! :) You can model this completely, i.e. describe height as a function of time. You can know exactly where the ball is at any particular time, and it follows a continuous curve. So it does make sense as a model, although it is perhaps a bit mind-bending.
- cperciva 16y agoYou don't give up, do you? An infinite series can have a finite sum. Yes, the ball bounces infinitely often, but it does that in a finite amount of time.
- greenlblue 16y agoI'm not disagreeing with you. Yes, an infinite series can have a finite sum, no argument there but what I am arguing with is that people are saying the ball stops bouncing after time t = whatever. Everyone is confounding two things here, air time and bouncing. Yes, the air time is finite but the bouncing isn't. So you can't say it stops bouncing after time t = whatever if you calculate t = whatever by assuming infinitely many bounces which is what everyone is doing because they are summing a geometric series where the terms of the series represent the air time of each bounce.
- cperciva 16y agoThere is a time t such that after time t the height of the ball's bounces is zero. Personally, I call that "not bouncing".
- greenlblue 16y agoOk, now we are in agreement because my definition of "stops bouncing" was finitely many bounces.
- paulofisch 16y agoBzzt Still wrong. If you have infinite bounces, and I ask you "Which number bounce in the series happens at exactly time X?", there is a time for X for which you will not be able to give an answer. This is because there is a limit for the latest time at which bounces happen. Yes, even with infinite bounces.
- greenlblue 16y agoWhatever. I got it sorted out so go bother someone else.
- D_Alex 16y agoI suspect this is what the intended solution to the puzzle is... Now for a follow up: Instead of assuming no friction, assume no slippage. What happens now?