Explanation

(d) Let the body be projected upwards with velocity $u$ from top of tower. Taking vertical downward motion of boy form top of tower to ground, we have
$u = - u,a = g = 10ms^{- 2},s = 50m,t = 10$s
As $s = ut + \frac{1}{2}at^{2}$,
So, $50 = - u \times 10 + \frac{1}{2} \times 10 \times 10^{2}$
On solving $u = 45ms^{- 1}$
If $t_{1}$ and $t_{2}$ are the timings taken by the ball to reach points $A$ and $B$ respectively, then
$20 = 45t_{1} + \frac{1}{2} \times 10 \times t_{1}^{2}$
and $40 = - 45t_{2} + \frac{1}{2} \times 10 \times t_{2}^{2}$
On solving, we get $t_{1} = 9.4\ $s and $t_{2} = 9.8$s
Time taken to cover the distance $AB$
$= \left( t_{2} - t_{1} \right) = 9.8 = 9.4 = 0.4$s.