# A tennis ball is thrown from a height h above the ground. If the coefficient of restitution e, what height will the ball achieve after the third collision ??

Dec 15, 2017

${h}_{3} = {e}^{6} {h}_{0}$, where ${h}_{0}$ is the initial height from which the ball was dropped and ${h}_{3}$ is the height the ball gains after the third collision.

#### Explanation:

Coefficient of restitution is defined as the ratio of relative speeds before and after collision.

This can also be expressed as the square-root of the ratio of kinetic energies before and after collision.

$e \setminus \equiv \setminus \frac{{v}_{p o s t}}{{v}_{p r e}} = \setminus \sqrt{\setminus \frac{{K}_{p o s t}}{{K}_{p r e}}}$.

Suppose if ${K}_{0}$ is the kinetic energy before the first collision and ${K}_{1} , {K}_{2}$ and ${K}_{3}$ are kinetic energies after the first, second and third collisions respectively,
$e = \setminus \sqrt{\setminus \frac{{K}_{1}}{{K}_{0}}} = \setminus \sqrt{\setminus \frac{{K}_{2}}{{K}_{1}}} = \setminus \sqrt{\setminus \frac{{K}_{3}}{{K}_{2}}}$

K_1 = e^2K_0; \qquad K_2 = e^2K_1=e^4K_0; \qquad K_3 = e^2K_2=e^6K_0

The height $h$ that a ball will ascend when thrown with a kinetic energy $K$ is found by applying the mechanical energy conservation condition,

mgh = K; \qquad \rightarrow \qquad h = K/(mg);

The ball is initially dropped from the height ${h}_{0}$ and acquires a kinetic energy ${K}_{0}$ when it hits the ground, before the first collision.

mgh_0 = K_0;

If ${h}_{1} , {h}_{2}$ and ${h}_{3}$ are the heights after the first, second and third collisions,

h_1 = K_1/(mg) = e^2 K_0/(mg) = e^2 (cancel{mg}h_0)/(cancel{mg}) = e^2h_0;

h_2 = K_2/(mg) = e^4K_0/(mg) = e^4 (cancel{mg}h_0)/(cancel{mg}) = e^4h_0;

h_3 = K_3/(mg) = e^6K_0/(mg) = e^6 (cancel{mg}h_0)/(cancel{mg}) = e^6h_0;

In general, the height gained by the ball after the ${n}^{t h}$ collision is:

${h}_{n} = {e}^{2 n} {h}_{0}$