Explanation

(b) Let $u$ be the initial upward velocity of the ball from $A$ and let $h$ be the height of the tower.
Taking the downward motion of the first stone from $A$ to the ground, we have
$h = - ut_{1} + \frac{1}{2}gt_{1}^{2}$ …(i)
Taking the downward motion of the second stone from $A$ to the ground, we have
$h = ut_{2} + \frac{1}{2}gt_{2}^{2}$ …(ii)
Multiplying Eq.(i) $t_{2}$ and Eq. (ii) by $t_{1}$ and adding we get
$h\left( t_{1} + t_{2} \right) = \frac{1}{2}gt_{1}t_{2}(t_{1} + t_{2})$
So, $h = \frac{1}{2}gt_{1}t_{2}$ …(iii)
For falls under gravity from the top of the tower
$h = \frac{1}{2}gt_{3}^{2}$ …(iv)
From Eqs. (iii) and (iv), $t_{3}^{2} = t_{1}t_{2}$ or $t_{3} = \sqrt{t_{1}t_{2}}$
$= \sqrt{6 \times 4}\ $= 6 s