Speed after n rebounds ?

Feb 27, 2017

${h}_{n} = \frac{{v}_{0}^{2} {\epsilon}^{2 n}}{2 g}$

Explanation:

After $n$ rebounds the ascending speed is

${v}_{n} = {v}_{0} {\epsilon}^{n}$ where ${v}_{0} = \sqrt{\frac{g h}{2}}$ is the initial speed.

with that initial speed, using the mechanical energy conservation relationship

$\frac{1}{2} m {v}_{n}^{2} = m g {h}_{n}$ or ${h}_{n} = {v}_{n}^{2} / \left(2 g\right) = \frac{{v}_{0}^{2} {\epsilon}^{2 n}}{2 g}$

then the equation for ${h}_{n}$ is

${h}_{n} = \frac{{v}_{0}^{2} {\epsilon}^{2 n}}{2 g}$