A particle of mass m is dropped from a height h above the ground. Simultaneously another particle of the…

The time for collision

$t_1=\frac{h}{\sqrt{2gh}}$

After $t_1$,

$v_{\mathrm{A}}=0-g{t}_1=-\sqrt{gh/2}$

$V_B=\sqrt{2gh}-g{t}_1=\sqrt{gh}\left[\sqrt{2}-\frac{1}{\sqrt{2}}\right]$

At the time of collision 

$\begin{array}{l}{\overrightarrow{P}}_i={\overrightarrow{P}}_f\\ \Rightarrow m{\overrightarrow{V}}_A+m{\overrightarrow{V}}_B=2m{\overrightarrow{V}}_f\\ \Rightarrow -\sqrt{\frac{gh}{2}}+\sqrt{gh}\left[\sqrt{2}-\frac{1}{\sqrt{2}}\right]=2{\overrightarrow{V}}_f\\ V_f=0\end{array}$

and height from ground

$=h-\frac{1}{2}g{t}_1^2=h-\frac{h}{4}=\frac{3h}{4}$

so time $=\sqrt{2\times \frac{\left(\frac{3h}{4}\right)}{g}}=\sqrt{\frac{3h}{2g}}$