Explanation

(d) Let the body after time $t/2$ be at $x$from the top, then
$x = \frac{1}{2}g\frac{t^{2}}{4} = \frac{gt^{2}}{8}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(i)$
$h = \frac{1}{2\ }gt^{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(ii)$
Eliminate $t$ from (i) and (ii), we get $x = \frac{h}{4}$
$\therefore$ Height of the body from the ground $= h - \frac{h}{4} = \frac{3h}{4}$