A particle of mass m is dropped from a height h above the ground simultaneously another particle of the same mass is thrown vertically upwards from the ground with a speed $\sqrt{2 g h} .$ If they collide head on completely inelastically, then the time taken for the combined mass to reach the ground is

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$\sqrt{\frac{h}{4 g}}$

$\sqrt{\frac{3 h}{4 g}}$

$\sqrt{\frac{3 h}{2 g}}$

$\sqrt{\frac{h}{2 g}}$

answer is B.

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Detailed Solution

The time for collision

$t_{1} = \frac{h}{v} = \frac{h}{\sqrt{2 g h}}$

After $t_{1},$

$v_{A} = 0 - g t_{1} = - \sqrt{g h / 2}$

$v_{B} = \sqrt{2 g h} - g t_{1} = \sqrt{g h} [\sqrt{2} - \frac{1}{\sqrt{2}}]$

At the time of collision $P_{i} = P_{f}$

$m \bar{V_{A}} + m \bar{V_{B}} = 2 m V_{f}$

$- \sqrt{\frac{g h}{2}} + \sqrt{g h} (\sqrt{2} - \frac{1}{2}) = 2 V_{f}$

$V_{f} = 0$ and height from ground $= h - \frac{1}{2} g t_{1}^{2}$

$H = h - \frac{h}{4} = \frac{3 h}{4}$

So time $= \sqrt{\frac{2 H}{g}} = \sqrt{\frac{3 h}{2 g}}$

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