Let body reaches the ground in $t$ seconds.
$\therefore $ Velocity of body after $(t-2)$ seconds from equation of motion
$v=u+g t'$
where $v$ is final velocity, $u$ the initial velocity
$(=0)$ and $t'=t-2$
$\therefore \, v=g (t-2)$
Distance covered in last two seconds
$h'=g(t-2) \times 2+\frac{1}{2} g(2)^{2} $
$60=20(t-2)+20$
On solving $t=4 \,s$
Hence, height of tower is given by
$h=u t+\frac{1}{2} g t^{2}$
Since, $ u=0$
$\therefore \,h =\frac{1}{2} g t^{2} $
$=\frac{1}{2} \times 10 \times(4)^{2}=80\, m $
