A body freely falling from a height h describes$\frac{\text{7h}}{\text{16}}$  in the last second of its fall. The height h is $\left(g=10{\text{ms}}^{-2}\right)$

Let ‘t’ be total time of fall

$\text{h}=\frac{1}{2}{\text{gt}}^{\text{2}}\to \left(1\right)$

In last second it travels$\frac{7\text{h}}{16}$  distance

In $\left(\text{t}-1\right)$ seconds

$\text{h}-\frac{\text{7h}}{\text{16}}=\frac{1}{2}\text{g}{\left(\text{t}-1\right)}^2$

$\frac{\text{9h}}{\text{16}}=\frac{1}{2}\text{g}{\left(\text{t}-1\right)}^2\to \left(2\right)$

$\frac{\left(2\right)}{\left(1\right)}=\frac{\frac{\text{9h}}{\text{16}}}{\text{h}}=\frac{\frac{1}{2}\text{g}{\left(\text{t}-1\right)}^2}{\frac{1}{2}{\text{gt}}^2}$

$\frac{\text{9h}}{\text{16}}={\left(\frac{\text{t}-1}{\text{t}}\right)}^2$

$\frac{\text{t}-1}{\text{t}}=\frac{3}{4}$

$1-\frac{\text{1}}{\text{t}}=\frac{3}{4}$

$1-\frac{3}{4}=\frac{\text{1}}{\text{t}}\Rightarrow \frac{\text{1}}{\text{t}}=\frac{1}{4}$

$\text{t}=4\sec$

$\text{h}=\frac{1}{2}{\text{gt}}^2$

$=\frac{1}{2}\left(10\right){\left(4\right)}^2$

$=5\times 16=80\text{m}$

$\text{h}=\text{80m}$