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 = 0.36 h

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

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

From (i) and (ii) we get,

t=5sec and  h=125m.