Explanation

(c) From equation of motion, we have
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ s = ut + \frac{1}{2}gt^{2}$
Where, $u$ is initial velocity, $g$ the acceleration due to gravity and $t$ the time.
For upward motion
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ h = - ut_{1} - \frac{1}{2}gt_{1}^{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(i)$
for downward motion
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ h = - ut_{2} + \frac{1}{2}gt_{2}^{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(ii)$
multiplying Eq. (i) by $t_{2}$ and Eq. (ii) by $t_{1}$ and subtracting Eq. (ii) by Eq. (i), we get
$\ \ \ h\left( t_{2} - t_{1} \right) = \frac{1}{2}gt_{1}t_{2}\left( t_{2} - t_{1} \right)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ h = \frac{1}{2}\ gt_{1}t_{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(iii)$
When stone is dropped from rest $u = 0,$ reaches the ground in $t$ second.
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ h = \frac{1}{2}gt^{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(iv)$
Equating Eqs. (iii) and (iv), we get
$\ \ \ \ \ \ \ \ \frac{1}{2}gt^{2} = \frac{1}{2}g\ t_{1}t_{2}$
$\Rightarrow \ \ \ \ \ \ \ \ t^{2} = t_{1}t_{2}\ \Rightarrow t = \sqrt{t_{1}t_{2}}$