A stone is dropped from a height $h$. Another stone is thrown up simultaneously from the ground, which can reach to maximum height of $4h$. The two stones will cross each other after time

The correct option is A $\sqrt{\frac{h}{8g}}$
Initial velocity of $2^{nd}$ stone.
Using $v^2=u^2+2as$
$0=u^2-2g\left(4h\right)$
$\Rightarrow u_2=\sqrt{8gh}$
Initial velocity of first stone is $u_1=0$
Relative velocity $u_{21}=u_2-u_1=\sqrt{8gh}$
Initial relative distance between stones is $d_{rel}=h$
Relative acceleration between stones is $a_{rel}=-g-(-g)=0$
Using $t=\frac{d_{rel}}{u_{rel}}$
$t=\frac{h}{\sqrt{8gh}}=\sqrt{\frac{h}{8g}}$