A man of $50 \mathrm{~kg}$ mass is standing in a gravity free space at a height of $10 \mathrm{~m}$ above…
A man of $50 \mathrm{~kg}$ mass is standing in a gravity free space at a height of $10 \mathrm{~m}$ above the floor. He throws a stone of $0.5 \mathrm{~kg}$ mass downwards with a speed $2 \mathrm{~ms}^{-1}$. When the stone reaches the floor, the distance of the man above the floor will be
$9.9 \mathrm{~m}$
$10.1 \mathrm{~m}$
$10 \mathrm{~m}$
$20 \mathrm{~m}$
Solution
$m r=$ constant
$\begin{aligned}
m_1 r_1 & =m_2 r_2 \\
r_2 & =\frac{m_1 r_1}{m_2} \\
& =\frac{0.5 \times 10}{50}=0.1
\end{aligned}$
The distance of the man above the floor (total height $)=10+0.1=10.1 \mathrm{~m}$.