Question

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

Answer 1

answer: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.