Question

In a gravity free space, a man of mass $M$ standing at a height $h$ above the floor, throws a ball of mass $m$ straight down with a speed $u$. When the ball reaches the floor, the distance of the man above the floor will be

• A. $h(1+\frac{m}{M})$
• B. $(1+\frac{M}{m})h$
• C. $h$
• D. $\frac{m}{M}h$

Answer 1

The correct option is A $h(1+\frac{m}{M})$

Applying Momentum Conservation
$MV+mu\left(\hat{j}\right)=0$
$V=\frac{mu}{M}(+\hat{j})$
Time taken by ball to reach ground $\Rightarrow h=ut$
$t=\frac{h}{u}$ (As $q=0$)
Therefore, distance covered by man $=vt=\frac{mh}{M}$
$\therefore$ Total height $=h+\frac{mh}{M}=h(1+\frac{m}{M})$