A particle of mass m is kept on a smooth cube of mass M and side L as shown in figure. Cube starts moving with a constant velocity v . Displacement of the center of mass along the horizontal direction when particle hits the ground is
 

Time taken by particle to hit the ground, $t=t_1+t_2$ here $t_1$ is the time to leave the block by the particle, and $t_2$ is the time to hit the ground after leaving the block.
Using $2^{\text{nd}}$ equation of motion
$\begin{array}{l}{t}_1=\frac{L}{v},\\ Also, h=\frac{1}{2}g{t}_2^2\\ \Rightarrow t_2=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2L}{g}}\\ \therefore t=\frac{L}{v}+\sqrt{\frac{2L}{g}}\end{array}$

Now, in horizontal direction, $V_{cm}=\frac{Mv}{m+M}$
The distance moved by the center of mass in this time,
$x_{cm}=v_{cm}t$
$\therefore x_{cm}=\frac{Mv}{m+M}\left[\frac{L}{v}+\sqrt{\frac{2L}{g}}\right]$
Therefore, the correct answer is (C)