A stone of mass 3kg fall from the top of a tower 100m high and buries itself 2m deep in the sand. The time of penetration is $\left(\mathrm{g}=10 \mathrm{m}/{\mathrm{s}}^2\right)$

Velocity before reaching ground
$\mathrm{v}=\sqrt{2\mathrm{gh}}=\sqrt{2\times 10\times 100}=20\sqrt{5}$
Acceleration of body penetrating
$\begin{array}{l}{\mathrm{v}}^2-{\mathrm{u}}^2=2\mathrm{as}\\ \mathrm{s}=2\mathrm{m},\mathrm{u}=20\sqrt{5}\mathrm{m}/\mathrm{s}\\ \mathrm{a}=\frac{{\mathrm{v}}^2-{\mathrm{u}}^2}{2\mathrm{s}}\\ =\frac{\left(20\right(\sqrt{5}){)}^2}{2\times 2}=\frac{2\times 10\times 100}{4}=500\\ \mathrm{a}=-500\mathrm{m}/{\mathrm{s}}^2\end{array}$
Time taken $\mathrm{v}=\mathrm{u}+\mathrm{at}$
$\begin{array}{l}\Rightarrow \mathrm{o}=\mathrm{u}-\mathrm{at}\Rightarrow \mathrm{u}=\mathrm{at}\\ \therefore \mathrm{t}=\frac{\mathrm{u}}{\mathrm{a}}=\frac{20\sqrt{5}}{500}=\frac{\sqrt{5}}{25}\sec \end{array}$