A body released from the top of a tower of height h takes T seconds to reach the ground. The position of the body at$\text{T/4}$ seconds is

$\text{t}=\frac{\text{T}}{\text{4}},\text{a}=\text{g}$

${\text{h}}^1=\text{ut}+\frac{1}{2}{\text{at}}^{\text{2}}$

$=\left(0\right)\left(\frac{\text{T}}{\text{4}}\right)+\frac{1}{2}g{\left(\frac{\text{T}}{\text{4}}\right)}^2$             

$=\frac{\text{g}{\text{.T}}^{\text{2}}}{2\times 16}$

$=\frac{\text{g}\text{.}\frac{\text{2h}}{\text{g}}}{\text{2×16}}$

$=\frac{\text{h}}{\text{16}}\text{m}$ from top of tower

$\text{=}\left(\text{h}-\frac{\text{h}}{\text{16}}\right)$ from ground

$=\frac{\text{15h}}{\text{16}}$ from ground