A body is thrown upward at an angle $\theta$ with horizontal from a point A on top of a tower. It reaches ground in time $t_1$. If it is thrown downward at an angle $\theta$ with horizontal from same point with same speed, it reaches the ground in time
$t_2$. If it is allowed to fall freely from A, then the time it takes to reach the ground is given by

The correct option is C $\sqrt{t_1{t}_2}$
If velocity of the body is $u_0$, then vertical component of velocity will be $u_0\sin \theta$ .
Sign convention - Taking upward direction from point A as positive .
When ball is thrown upward $-h=\left(u_0\sin \theta \right)t_1-\frac{g{t_1}^2}{2}$ $\to$(i)
When ball is thrown downward $-h=-\left(u_0\sin \theta \right)t_2-\frac{g{t_2}^2}{2}$$\to$(ii)
Subtracting (ii) from (i) , we get $0=\left(u_0\sin \theta \right)(t_1+t_2)+\frac{g\left({t_2}^2-{t_1}^2\right)}{2}$ $\Rightarrow u_0\sin \theta =\frac{g(t_1-t_2)}{2}$.
Putting above in equation (ii), we get $-h=-\frac{g(t_1-t_2)t_2}{2}-\frac{g{t_2}^2}{2}\Rightarrow h=\frac{g{t}_1{t}_2}{2}$ .
For free fall initial velocity = 0 $\Rightarrow h=\frac{g{t}^2}{2}$ .
Equating h, we get $\frac{g{t}_1{t}_2}{2}=\frac{g{t}^2}{2}\Rightarrow t=\sqrt{t_1{t}_2}$