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 (g=10 ms−2 )

Given that

A body is falling freely from height = h

It has covered 36% of h in the time last 1s

So we get $t_{1} = (t - 1)$ for descending through 64% h

$h = ϑ t + \frac{1}{2} {gt}^{2} \dots \dots (1)$

$64 % h = \frac{1}{2} g {(t - 1)}^{2} \dots \dots (2)$

$\frac{(1)}{(2)} = \frac{h}{\frac{64}{100} \times h} = \frac{\frac{1}{2} {gt}^{2}}{\frac{1}{2} g {(t - 1)}^{2}}$

$\frac{100}{64} = \frac{t^{2}}{{(t - 1)}^{2}} = \frac{10}{8} = \frac{t}{t - 1}$

$8 t = 10 t - 10$

$2 t = 10 \Rightarrow t = 5 s$

$h = \frac{1}{2} {at}^{2} = \frac{1}{2} \times 10 \times 25 = 125 m$

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 (g= $10 {ms}^{- 2}$ )