When dropped on a hard surface, a rubber ball takes a series of bounces, each one about 4/5 as high as the preceding one. If the rubber ball is dropped from a height of 20 ft, what is the approximate distance the ball travels before coming to rest? Please explain how you got the answer. Thanks!

Answer 1

assume an infinite number of hops. In practice, of course, the ball stops earlier than that, but you have a geometric progression where

a = 20
r = 4/5

So, the initial drop is 20, and each bounce is a round trip 4/5 as high as the one before, so the total distance after n hops is

20 + 2(20 * 4/5) + 2(20*4/5)^2 + ...

So, the real GP is

a = 40
r = 4/5
and 20 must be subtracted because the first hop did not have to go up.

The total distance is thus

40/(1 - 4/5) - 20 = 180 ft

Answer 2

Thank you so much!