A stone is released from a height H  reaches  the ground in n seconds. The height of stone  after $\frac{n}{2}$ seconds is

given time of descent =t== n;   height from the ground=h =H

$$\begin{gathered} we know, s=ut + \frac{1}{2}a{t}^2 \\ substitute s=h; a=g=acceleration due to gravity; \\ time of descent=t=\sqrt{\frac{2h}{g}} \\ substitute t=n; h=H \end{gathered}$$

    $$\begin{gathered} n=\sqrt{\frac{2H}{g}} \\ \Rightarrow n \alpha \sqrt{H} \\ H=\frac{1}{2}g{n}^2-\left(1\right) \end{gathered}$$
Distance travelled in $\frac{n}{2}$ sec is
$$\begin{gathered} we know, for freely falling body,u=0 and displacement=S=\frac{1}{2}g{\left(t\right)}^2 \\ S=\frac{1}{2}g{\left(\frac{n}{2}\right)}^2 \\ \Rightarrow S=\frac{\frac{1}{2}g{n}^2}{4} \\ substitute eqution \left(1\right) \\ \Rightarrow S=\frac{H}{4}---\left(2\right) \end{gathered}$$ 
The height from the ground 
$\begin{array}{l}H\text{'}=H-S\\ substitute equation \left(2\right)\\ H\text{'}=H-\frac{H}{4}\\ \Rightarrow H\text{'}=\frac{3H}{4}=height from the ground\end{array}$