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=$10{\text{ms}}^{-2}$ )

A body is falling freely from height = h

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

So we get ${\text{t}}_{\text{1}}=\left(\text{t}-1\right)$  for descending through 64% h

$\text{h}=\vartheta t+\frac{1}{2}{\text{gt}}^{\text{2}}\cdots \cdots \left(1\right)$

$64\%\text{h}=\frac{1}{2}\text{g}{\left(\text{t}-1\right)}^2\cdots \cdots \left(2\right)$   

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

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

$8\text{t}=10\text{t}-10$

$2\text{t}=10\Rightarrow \text{t}=5\text{s}$

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