Thank you for reporting, we will resolve it shortly
Q.
A body ‘ $A$’ is dropped vertically from the top of a tower. If another identical body ‘$B$’ is projected from the same point at the same instant, then
As the body ' $A$ ' is dropped from rest
$\therefore h=\frac{1}{2} g t_{A}^{2} $
$\Rightarrow t_{A}=\sqrt{\frac{2 h}{g}}$
As the body $B$ is given a horizontal velocity at the time of release, it is going to follow the same trajectory as a body on a projectile motion projected with a velocity having the same horizontal component as the horizontal velocity of $B$ given at $H$. So the time taken by $B$ from $H$ to $Y$ is same as that from $X$ to $H$.
Now let the vertical component of velocity at $X$ be $v$ then at $H$
$0=v^{2}-2 g h \,\,\,\,\,\,\left[\therefore a t H, v_{f}=0\right]$
$\Rightarrow v^{2}=2 g h $
$\Rightarrow v=\sqrt{2 g h}$
and $0=v-g t$
$\Rightarrow t=\frac{v}{g}=\frac{\sqrt{2 g h}}{g}=\sqrt{\frac{2 h}{g}}$
$\therefore t_{B}=t=\sqrt{\frac{2 h}{g}}=t_{A}$
