A body falls freely from the top of a tower. It covers 36% of the total height in the last second before striking the ground level. The height of the tower is 

Let height of tower is h and body takes t time to reach to ground when it fall freely.

$\therefore h=\frac{1}{2}g{t}^2 .....\left(i\right)$

In last second, i.e., t^th sec body travels=0.36h
It means in rest of the time i.e. in (t - 1) sec it travels

$=h-0.36h=0.64h$

Now applying equation of motion for (t - 1) sec, we get

$0.64h=\frac{1}{2}g(t-1{)}^2 ......(ii)$

From (i) and (ii), we get t=5 sec and h=125 m.