Explanation

(d) Let height of tower is $h$ and body takes $t$ time to reach to ground when it fall freely
$\therefore h = \frac{1}{2}gt^{2}$
In last second i.e. $t^{th}\ sec$ body travels =$0.36\ h$
It means in rest of the time i.e. in $(t - 1)sec$ it travels
$= h - 0.36\ h = 0.64\ h$
Now applying equation of motion for $(t - 1)sec$
$0.64\ h = \frac{1}{2}g(t - t)^{2}$
From (i) and (ii) we get,$t = 5\sec\ $and $h = 125m$