**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#