Explanation

(c) Let the stone falls through a height $h$ in $t\ s$
Here, $u = 0,\ a = g$
Using $D_{n} = u + \frac{a}{2}(2n - 1)$
Distance travelled by the stone in the last second is
$\frac{9h}{25} = \frac{g}{2}(2t - 1)\ \ \ \lbrack\because u = 0\rbrack$ …(i)
Distance travelled by the stone in $t\ s$ is
$h = \frac{1}{2}\ gt^{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \lbrack\because u = 0\rbrack$ …(ii)
Divide (i) by (ii), we get
$\frac{9}{25} = \frac{(2t - 1)}{t^{2}}$
$9t^{2} = 50t - 25$
$9t^{2} - 50t + 25 = 0$
On solving, we get
$t = 5s$ or $t = \frac{5}{9}s$
Substituting $t = 5s$ in (ii), we get
$h = \frac{1}{2} \times 9.8 \times (5)^{2} = 122.5\ m$