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 $\sqrt{2gh}$. If they collide head-on completely inelastically, the time taken for the combined mass to reach the ground, in units of $\sqrt{\dfrac{h}{g}}$ is:
Answer: (D) $\sqrt{\dfrac{3}{2}}$
Relative to each other the particles move with constant speed $\sqrt{2gh}$ (both have the same acceleration $g$), so they meet after
$$t = \frac{h}{\sqrt{2gh}} = \sqrt{\frac{h}{2g}}$$
At that moment: falling particle $v_1 = gt = \sqrt{gh/2}$ (down); rising particle $v_2 = \sqrt{2gh} - gt = \sqrt{gh/2}$ (up).
Equal masses with equal and opposite velocities: after sticking, the combined mass is momentarily at rest.
Height of collision $= h - \tfrac{1}{2}gt^2 = h - \tfrac{h}{4} = \tfrac{3h}{4}$.
Time to fall from rest: $\sqrt{\dfrac{2(3h/4)}{g}} = \sqrt{\dfrac{3}{2}}\sqrt{\dfrac{h}{g}}$.
Solution by Sreeraj P, M.Sc Physics