Asked by Ashley
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!
Answers
Answered by
Steve
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
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
Answered by
Ashley
Thank you so much!