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 ??

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 \equiv \frac{v_{post}}{v_{pre}} = \sqrt{\frac{K_{post}}{K_{pre}}}#.

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=\sqrt{\frac{K_1}{K_0}} = \sqrt{\frac{K_2}{K_1}} = \sqrt{\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^{th}# collision is:

#h_n = e^{2n}h_0#