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(D) For the body of mass $m$ falling from height $h$,the time $t$ taken to reach height $h/2$ is given by: $\frac{h}{2} = \frac{1}{2}gt^2 \Rightarrow

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$\frac{\sqrt{10gh}}{3}$

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(D) For the body of mass $m$ falling from height $h$,the time $t$ taken to reach height $h/2$ is given by: $\frac{h}{2} = \frac{1}{2}gt^2 \Rightarrow t = \sqrt{\frac{h}{g}}$.For the body of mass $2m$ thrown upwards with velocity $v$,the displacement at time $t$ is: $\frac{h}{2} = vt - \frac{1}{2}gt^2$. Substituting $t = \sqrt{h/g}$,we get $\frac{h}{2} = v\sqrt{\frac{h}{g}} - \frac{h}{2} \Rightarrow v = \sqrt{gh}$.At the moment of collision at $h/2$,the velocity of mass $m$ is $v_1 = -gt = -\sqrt{gh}$ (downward) and the velocity of mass $2m$ is $v_2 = v - gt = \sqrt{gh} - \sqrt{gh} = 0$.By conservation of momentum: $m(-v_1) + 2m(v_2) = (m + 2m)V_{combined} \Rightarrow m(-\sqrt{gh}) + 2m(0) = 3mV_{combined} \Rightarrow V_{combined} = -\frac{1}{3}\sqrt{gh}$.The combined mass $3m$ is at height $h/2$ moving downwards with speed $u = \frac{1}{3}\sqrt{gh}$.Using $v_f^2 = u^2 + 2g(h/2)$,we get $v_f^2 = (\frac{1}{9}gh) + gh = \frac{10}{9}gh$.Thus,$v_f = \frac{\sqrt{10gh}}{3}$.

$A$ body of mass $m$ is dropped from a height of $h$. Simultaneously,another body of mass $2m$ is thrown vertically upward with such a velocity $v$ that they collide at a height $h/2$. If the collision is perfectly inelastic,the velocity at the time of collision with the ground will be:

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