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 $\frac{v^2}{2g}=4h$
$\Rightarrow u^2=8gh$
$\therefore u=\sqrt{8gh}$

Now,
$h_1=\frac{1}{2}g{t}^2$
$h_2=\sqrt{8ght}-\frac{1}{2}g{t}^2$
where, $t=$ time to cross each other.
$\therefore h_1+h_2=h$
$\Rightarrow \frac{1}{2}g{t}^2+\sqrt{8ght}-\frac{1}{2}g{t}^2=h$
$\Rightarrow t=\frac{h}{\sqrt{8gh}}$
$=\sqrt{\left(\frac{h}{8g}\right)}$