A glass marble dropped from a certain height above the horizontal surface reaches the surface in time t and then continues to bounce up and down. The time in which the marble finally comes to rest is

$$\begin{gathered} {\mathrm{t}}_{\mathrm{AB}}=\sqrt{\frac{2\mathrm{h}}{\mathrm{g}}} \\ {t}_{BC}+t_{CB}=2\sqrt{\frac{2h_1}{g}} \\ =2\sqrt{\frac{2e^{2}h}{g}}=2e\sqrt{\frac{2h}{g}} \\ {t}_{BD}+t_{DB}=2e^2\sqrt{\frac{2h}{g}} \end{gathered}$$

$\therefore$ Total time taken by the body in coming to rest

$$\begin{gathered} =\sqrt{\frac{2h}{g}}+2e\sqrt{\frac{2h}{g}}+2e^2\sqrt{\frac{2h}{g}}+......... \\ =\sqrt{\frac{2h}{g}}+2e\sqrt{\frac{2h}{g}}\left[1+e+e^2+.........\right] \\ =\sqrt{\frac{2h}{g}}+2e\sqrt{\frac{2h}{g}}\times \frac{1}{1-e}=\sqrt{\frac{2h}{g}}\left[\frac{1+e}{1-e}\right]=t\left(\frac{1+e}{1-e}\right) \end{gathered}$$