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

time for collision ${t}_{1} = \dfrac{h}{\sqrt{2gh}}$
After ${t}_{1}$
${V}_{A} = 0 - g{t}_{1} = - \sqrt{\dfrac{gh}{2}}$
and ${V}_{B} = \sqrt{2gh} - g{t}_{1} = \sqrt{gh}\left[\sqrt{2} - \dfrac{1}{\sqrt{2}}\right]$
at the time of collision
${\overset{\to}{p}}_{i} = \overset{\to}{{p}_{f}}$
$⇒ m{\overset{\to}{V}}_{A} + m{\overset{\to}{V}}_{B} = 2m{\overset{\to}{V}}_{f}$
$⇒ - \sqrt{\dfrac{gh}{2}} + \sqrt{gh}\left[\sqrt{2} - \dfrac{1}{\sqrt{2}}\right] = 2{\overset{\to}{V}}_{f}$
${V}_{f} = 0$

and height from ground

$= h - \dfrac{1}{2}g{t}_{1}^{2} = h - \dfrac{h}{4} = \dfrac{3h}{4}$

so time ,

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