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 $h(1+\frac{m}{M})$

Centre of mass will remain at height h.
$\therefore h_{CM}=\frac{m\times 0+MH}{m+M}=h$
$\therefore H=h(1+\frac{m}{M})$