Guest Posted April 22, 2008 Report Share Posted April 22, 2008 A ball is dropped from a distance of 11m. Each time, it bounces back to a height which is half of the previous one. What is the total distance travelled by the ball when it comes to rest. Quote Link to comment Share on other sites More sharing options...
0 Brandonb Posted April 22, 2008 Report Share Posted April 22, 2008 It never comes to a rest. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 22, 2008 Report Share Posted April 22, 2008 A ball is dropped from a distance of 11m. Each time, it bounces back to a height which is half of the previous one. What is the total distance travelled by the ball when it comes to rest. The ball would never come to a rest, since the height of the bounce is always greater than zero. I suppose limits could be used to determine the distance to be traveled, but it's been too long to remember those. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 22, 2008 Report Share Posted April 22, 2008 It never comes to a rest. Sorry, I forgot to mention . What I require is the approximate distance . Quote Link to comment Share on other sites More sharing options...
0 Brandonb Posted April 22, 2008 Report Share Posted April 22, 2008 Sorry, I forgot to mention . What I require is the approximate distance . Easy The answer is INFINITE. No approximations on this, it goes... forever. Infinite distance. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 22, 2008 Report Share Posted April 22, 2008 Sorry, I forgot to mention . What I require is the approximate distance . ~~~~22m Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 22, 2008 Report Share Posted April 22, 2008 I'm the king at posting 1 minute after Brandonb. Too fast, too fast... Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 22, 2008 Report Share Posted April 22, 2008 Easy The answer is INFINITE. No approximations on this, it goes... forever. Infinite distance. Sorry Brandonb, but that is not the correct answer. You see, the distance will keep on decreasing. Just think what will happen if it touches 1. Quote Link to comment Share on other sites More sharing options...
0 Brandonb Posted April 22, 2008 Report Share Posted April 22, 2008 it's almost 11m Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 22, 2008 Report Share Posted April 22, 2008 I'm the king at posting 1 minute after Brandonb. Too fast, too fast... Are you both having a telephone conversation? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 22, 2008 Report Share Posted April 22, 2008 Well, if we consider we are not all mathematicians, and love our calculus, then we need to become physicists. In which case, a ball would need to compress for it to be able to gain the force to rebound. And at about 22m, the size of the ball becomes relevant, for a 1m diameter beach ball is going to travel further in total than a marble... What kind of material is the ball made of, and what is it bouncing upon? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 22, 2008 Report Share Posted April 22, 2008 ~~~~22m Good try But the answer is a bit greater . Try considering some more iterations . Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 22, 2008 Report Share Posted April 22, 2008 the distance is 11m. when the ball hits the ground, there is a very very very short time that it is at rest. think of it as a parabala with a minimum value of zero. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 22, 2008 Report Share Posted April 22, 2008 Sorry, I forgot to mention . What I require is the approximate distance . Why approximate? There is an exact answer to 1 + .5 + .25 + .125... = x Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 22, 2008 Report Share Posted April 22, 2008 the distance is 11m. when the ball hits the ground, there is a very very very short time that it is at rest. think of it as a parabala with a minimum value of zero. I would be so adventurous to define "at rest" as remaining still or not moving over some time interval. If you take the ball's motion out of context of time, then at any time T the ball would, by your definition, be at rest. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 22, 2008 Report Share Posted April 22, 2008 the distance is 11m. when the ball hits the ground, there is a very very very short time that it is at rest. think of it as a parabala with a minimum value of zero. Ok giterdone, I'll modify the question for you: What is the total distance travelled by the ball when it finally comes to rest. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 22, 2008 Report Share Posted April 22, 2008 Ok, assuming that it comes to a complete rest, then is it a ball anymore? (ie round) Once hitting the ground, it will compress, preparing for the bounce, and be somewhat flatter at the base. Does the base point travel less distance that the top....? (Assuming we are dealing with a beach ball here) If the initial 11m is measures from the core centre, then in could not fall a total of 11m... Quote Link to comment Share on other sites More sharing options...
0 Brandonb Posted April 22, 2008 Report Share Posted April 22, 2008 33 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 22, 2008 Report Share Posted April 22, 2008 I guess it's gotta be 33. Our equation is 11 + 2*11/2 + 2*11/4 + 2*11/8 ... etc which equals 11 + 2*Sum(11/2^n) for values of n from 1 to infinity. That simplifies to 11 + 22*Sum(1/2^n). As n approaches infinity, Sum(1/2^n) approaches 1. Therefore the distance is approximately 11 + 22*1 = 33. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 22, 2008 Report Share Posted April 22, 2008 Ok, assuming that it comes to a complete rest, then is it a ball anymore? (ie round) Yes, a ball that has come to rest is still defined as a ball, regardless of whether it's perfectly round (or ever was). Once hitting the ground, it will compress, preparing for the bounce, and be somewhat flatter at the base. Does the base point travel less distance that the top....? (Assuming we are dealing with a beach ball here) If the initial 11m is measures from the core centre, then in could not fall a total of 11m... Assume the ball traveled 11m on the first fall. Please read guideline #5 in the sticky titled "Important: READ BEFORE POSTING". Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 22, 2008 Report Share Posted April 22, 2008 That was a nice variation of Zeno's dichotomy paradox. The solution is a fairly simple limit applied to a sum of infinitely decreasing values, but this depends on the acceptance that this sum can be expressed as a finite limit. Of course, practical experience tells us it can, or else there could never be motion in the first place. Good problem. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 22, 2008 Report Share Posted April 22, 2008 Right. Got it. So the ball falls, bounces and travels 33 metres before coming to rest. Right? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 22, 2008 Report Share Posted April 22, 2008 I guess it's gotta be 33. Our equation is 11 + 2*11/2 + 2*11/4 + 2*11/8 ... etc which equals 11 + 2*Sum(11/2^n) for values of n from 1 to infinity. That simplifies to 11 + 22*Sum(1/2^n). As n approaches infinity, Sum(1/2^n) approaches 1. Therefore the distance is approximately 11 + 22*1 = 33. The equation is not 11 + 2*11/2 + 2*11/4 + 2*11/8 ... but it's 11 + 11/2 + 11/4 + 11/8 + ... This can be expressed as 11(1 + 1/2 + 1/4 + 1/8 + ... ) = 11(2) = 22 Still as we said before, the ball never comes to rest, so it should be worded a bit differently. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 22, 2008 Report Share Posted April 22, 2008 The equation is not 11 + 2*11/2 + 2*11/4 + 2*11/8 ... but it's 11 + 11/2 + 11/4 + 11/8 + ... This can be expressed as 11(1 + 1/2 + 1/4 + 1/8 + ... ) = 11(2) = 22 No, Artificial I and Aussie Bandit are right. Remember, it drops 11m on the first drop, but it bounces up 5.5m and then falls 5.5m. So it's 11 + 11 + 5.5... or the formulas that Artificial I provided. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 22, 2008 Report Share Posted April 22, 2008 No, Artificial I and Aussie Bandit are right. Remember, it drops 11m on the first drop, but it bounces up 5.5m and then falls 5.5m. So it's 11 + 11 + 5.5... or the formulas that Artificial I provided. Ah, of course. Forgot about the up AND down. Thanks. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 22, 2008 Report Share Posted April 22, 2008 33 Quote Link to comment Share on other sites More sharing options...
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A ball is dropped from a distance of 11m.
Each time, it bounces back to a height which is half of the previous one.
What is the total distance travelled by the ball when it comes to rest.
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