Explanation

(c) Let body reaches the ground in $t$ sec.
$\therefore$ Velocity of body after $(t - 2)$ sec from equation of motion.
$v = u + gt'$
And $t^{'} = t - 2\ $
$\therefore v = g(t - 2)$
Distance covered in last two sec
$h^{'} = g(t - 2) \times 2 + \frac{1}{2}g(2)^{2}$
$60 = 20(t - 2) + 20$
Or $t = 4\ s$
Hence, height of tower is given buy
$h = ut + \frac{1}{2}gt^{2}$
$h = \frac{1}{2}gt^{2}\lbrack\therefore u = 0\rbrack$
$\ \ \ = \frac{1}{2} \times 10 \times (4)^{2} = 80\ m.$