A ball is held at rest at some height above a hard, horizontal surface. Once the ball is released it falls, hits the surface, and starts bouncing vertically up and down. Suppose that with each bounce the ball loses a fixed fraction p (with 1>p>0) of its energy. This loss could be due to a number of reasons (inelasticity, drag, etc) that are left unspecified.How many times will the ball bounce before coming to rest (if at all)? Provide a detailed explanation of your reasoning, not simply a one-line answer.How long will it take for the ball to come to rest (if at all)? Give your answer as a formula that contains as variables only p and the time T1 from the moment that the ball was released to the first contact with the horizontal surface.

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GOURAVDAYMAGO GOURAVDAYMAGO · Answer 1

It can never theoretically become 0. After n bounces, it may be closer to 0 or extremely tiny, depending on the precise values of p and T1.

initial height dropped =h = gT12/2

T1 - time from the moment that the ball was released to the first contact with the horizontal surface.

initial energy T = mgh

after first bounce energy loss = mghp

height it will raise after first bounce mgh1 = mgh(1-p)

after each bounce its energy is reduced by p

after n bounces it will raise to a height

mghn=mgh(1-p)n

hn= h(1-p)n = gT12/2 *(1-p)n

gT₁² (1-p)n/2

Theoritically hn can never become 0. It can be closer to 0 or can be negligibly small after n bounces, depending on the actual values of p and T1 .

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