A particle of mass m is dropped from a height h above the ground. At the same time another particle of the same mass is thrown vertically upwards from the ground with a speed of $\sqrt{2\mathrm{gh}}$ . If they collide head – on completely inelastically. The time taken for the combined mass to reach the ground in units of $\sqrt{\frac{\mathrm{h}}{\mathrm{g}}}$is

Time of collision $\mathrm{t}=\frac{\mathrm{h}}{\sqrt{2\mathrm{gh}}}$

at the time of collision

${\mathrm{V}}_{\mathrm{A}}=-\mathrm{gt}=-\sqrt{\frac{\mathrm{gh}}{2}}$

and ${\mathrm{V}}_{\mathrm{B}}=\sqrt{2\mathrm{gh}}-\mathrm{gt}$

Apply common velocity $\mathrm{V}=\frac{{\mathrm{mV}}_{\mathrm{A}}+{\mathrm{mV}}_{\mathrm{B}}}{2\mathrm{m}}$

${\mathrm{V}}_{\mathrm{common}}=\frac{\mathrm{m}\left(-\sqrt{{\displaystyle \frac{\mathrm{gh}}{2}}}\right)+\mathrm{m}\left(\sqrt{2\mathrm{gh}}-\sqrt{\frac{\mathrm{gh}}{2}}\right)}{2\mathrm{m}}=0$

And the combined mass is at a height given by 

$\mathrm{h}-\frac{1}{2}\mathrm{g}{\left(\sqrt{\frac{\mathrm{h}}{2\mathrm{g}}}\right)}^2=\frac{3\mathrm{h}}{4}$(Object A was freely falling for $\sqrt{\frac{\mathrm{h}}{2\mathrm{g}}}$ seconds)

So, the time taken for the combined mass to freely fall through $\frac{3\mathrm{h}}{4}$is given by 

$\frac{3\mathrm{h}}{4}=\frac{1}{2}{\mathrm{gT}}^2\Rightarrow \mathrm{T}=\sqrt{\frac{3\mathrm{h}}{2\mathrm{g}}}$