A body released from certain height above the ground describes $\frac{7}{16}$ of the total height in the last second of its fall. Then it is falling from a height equal to $\left(g=9.8m/s^2\right)$

Let ‘n’ be the last second of its fall
 the total height covered after ‘n’ sec is
$\begin{array}{l}h=ut+\frac{1}{2}g{t}^2\\ \Rightarrow h=0+\frac{1}{2}g{n}^2\\ =\frac{1}{2}g{n}^2\dots \dots \dots ..\left(1\right)\end{array}$

The distance travelled in the last second of its fall is  $s_n=u+g\left(n-\frac{1}{2}\right)$

$\begin{array}{l}\Rightarrow \frac{7}{16}h=0+g\frac{(2n-1)}{2}\\ \Rightarrow \frac{7}{16}\left(\frac{1}{2}g{n}^2\right)=g\frac{(2n-1)}{2}\\ \Rightarrow \frac{2n-1}{n^2}=\frac{7}{16}\end{array}$

On solving  $n=4s$
$\therefore$ The time taken by the body  $t=4s$
$\therefore$ The height  $h=\frac{1}{2}g{n}^2$              

                         $\begin{array}{l}=\frac{1}{2}\times 9.8\times (4{)}^2\\ =9.8\times 8\\ =78.4\mathrm{m}\end{array}$