When a ball dropped from a height $h$, then it will strick the floor with velocity $v=\sqrt{2 g h}$ Now, after strick the floor it will bounce back with velocity, $\mathrm{u}=\mathrm{ev}$
Now it will travel $2 \mathrm{~h}_1$ distance before again striking the ground.
Here $h_1=\frac{e^2 v^2}{2 g}$ again the ball will bounce back with $e^2 v$ velocity.
This process continues for long time till ball comes to the rest.
So, total distance travelled $=\mathrm{h}+2 \mathrm{~h}_1+2 \mathrm{~h}_2+\ldots$
$$
\begin{aligned}
& =\frac{v^2}{2 g}+\frac{2 e^2 v^2}{2 g}+\frac{2 e^4 v^2}{2 g}+\ldots \\
& =\frac{v^2}{2 g}\left[1+2 e^2+2 e^4+2 e^6+\ldots\right] \\
& =\frac{v^2}{2 g}+\frac{2 e^2 v^2}{2 g}-\left[1+e^2+e^4+\ldots\right] \\
& =\frac{v^2}{2 g}+\frac{e^2 v^2}{g}\left(\frac{1}{1-e^2}\right) \\
& =\frac{v^2}{2 g}\left[\frac{1-e^2+2 e^2}{1-e^2}\right]=\frac{v^2}{2 g}\left(\frac{1+e^2}{1-e^2}\right)=h\left(\frac{1+e^2}{1-e^2}\right)
\end{aligned}
$$