A ball is released from height h w.r.t. the ground. Draw a qualitative height versus time diagram of the ball bouncing in a non-ideal case.

In a non-ideal case, there will be energy loss in heat when the ball touches the ground. In particular:Kn=aKn-1, where Kn-1 is the kinetic energy before the nth bounce, Kn is the kinetic energy after the bounce, and a is the fraction of kinetic energy that remains after the bounce. We can see that this produces a geometric series of the form: Kn=anK0, which gives the kinetic energy after n bounces. To transform this into height, we just need to remember that the maximum height after n bounces hn is reached when Kn is all converted into potential energy (mgh), where m is the mass of the ball. Substitute the formula of KE. Hence: hn=Kn/(mg)=anK0/(mg). Now, as K0 is proportional to the height to which the ball was originally released h0(again, for the conservation of energy), We get: hn \propto anh0. Hence the maximum height decrease exponentially (as a <1). The maximum vertices are also peaks of rotated parabolas, as the ball obeys the free-fall equation which says that h is proportional to t2. Draw this and you get the diagram requested.

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Answered by Emanuele P. Physics tutor

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