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

time for collision $t_1=\frac{h}{\sqrt{2gh}}$
After $t_1$
$V_A=0-g{t}_1=-\sqrt{\frac{gh}{2}}$
and $V_B=\sqrt{2gh}-g{t}_1=\sqrt{gh}\left[\sqrt{2}-\frac{1}{\sqrt{2}}\right]$
at the time of collision
${\overrightarrow{p}}_i=\overrightarrow{p_f}$
$\Rightarrow m{\overrightarrow{V}}_A+m{\overrightarrow{V}}_B=2m{\overrightarrow{V}}_f$
$\Rightarrow -\sqrt{\frac{gh}{2}}+\sqrt{gh}\left[\sqrt{2}-\frac{1}{\sqrt{2}}\right]=2{\overrightarrow{V}}_f$
$V_f=0$

and height from ground

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

so time ,

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