A superball is dropped from rest from a height of 2.0m. It bounces repeatedly from the floor, as superballs are prone to do. After each bounce the ball dissipates some energy, so eventually it comes to rest. The following pattern is observed: After the 1st bounce, the ball returns to a maximum height that is 3/4 of its initial height. After the 2nd bounce, the ball returns to a maximum height that is 3/4 of its maximum height after the 1st; After the 3rd bounce, the ball returns to a maximum height that is 3/4 of its maximum height after the second, etc. In fact, for this particular ball, the maximum height is achieved after the nth bounce is found to be 3/4 of the maximum height achieved in the previous bounce. If this pattern is repeated, how many times will the ball bounce before coming to rest, and how long the process will take? (Neglect air friction.)

Question

Steve · Answer

1st fall: 2m
nth bounce: 2*(3/4)^n

So, you have to decide what "at rest" means. At some point the friction will overcome the tendency to bounce, and the bounces will stop. In theory, the ball bounces forever, each time achieving less height.

So, once you decide how high the minimum acceptable bounce is, just solve for n.

For example, if the ball is considered at rest when the bounces are less than .01mm high, then

2(3/4)^n = .00001
(3/4)^n = .000005
n = (log .000005)/log(3/4) = 42.4 bounces

The time for the nth bounce is
t = √(2*(3/4)^n)/4.9) = √(2/4.9) √(3/4)^n

Figure the time for the 1st fall, then add up as many terms for the bounce times as you decide you need.

Take a peek at

http://www.wolframalpha.com/input/?i=sum+%E2%88%9A%282*%283%2F4%29^n%29%2F4.9%29