Question

A stone is dropped from a height $h$ simultaneously, another stone is thrown up from the ground which reaches a height $4h$ . The two stones cross each other after time

• A. $ \sqrt{\frac{h}{8g}} $
• B. $ \sqrt{8gh} $
• C. $ \sqrt{2gh} $
• D. $ \sqrt{\left( \frac{h}{2g} \right)} $

Answer 1

Answer: (A)

For first stone $v=0$ and
For second stone $\frac{v^{2}}{2 g}=4 h$
$\Rightarrow u^{2}=8 g h$
$\therefore u=\sqrt{8 g h}$

Now,
$h_{1}=\frac{1}{2} g t^{2}$
$h_{2}=\sqrt{8 g h t}-\frac{1}{2} g t^{2}$
where, $t=$ time to cross each other.
$\therefore h_{1}+h_{2}=h$
$\Rightarrow \frac{1}{2} g t^{2}+\sqrt{8 g h t}-\frac{1}{2} g t^{2}=h$
$\Rightarrow t=\frac{h}{\sqrt{8 g h}}$
$=\sqrt{\left(\frac{h}{8 g}\right)}$