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

The Correct Option is 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. $\therefore h_{1}+h_{2}=h$ $\Rightarrow \frac{1}{2} g t^{2}+\sqrt{8 g h t}-\frac{1}{2} g t^{2}=h$ $\Rightarrow t=\frac{h}{\sqrt{8 g h}}$ $=\sqrt{\left(\frac{h}{8 g}\right)}$