In a gravity free room a man of mass $m_1$ is standing at a height $h$ above the floor. He throws a ball of mass $m_2$ vertically downward with a speed $u$. Find the distance of the man from the floor when the ball reaches the ground.

Time taken by ball to reach the ground $t=\frac{h}{u}$

By conservation of linear momentum, speed of man $v=\left(\frac{m_{2}u}{m_1}\right)$

 Therefore, the man will move upward by a distance $=vt=\left(\frac{h}{u}\right)\left(\frac{m_{2}u}{m_1}\right)=\frac{m_2}{m_1}h$

Total distance of the man from the floor $=h+\frac{m_2}{m_1}h=\left(1+\frac{m_2}{m_1}\right)h$.