A stone weighing $3kg$ falls from the top of a tower $100m$ high and buries itself $2m$ deep in sand. Find the time (in seconds) of penetration $(g=10m/s^2)$

The correct option is D $\frac{\sqrt{5}}{25}$

Taking downward as positive,
For AB, $u=0;S=100m;a=g$
Using third equation of motion,
$v^2=u^2+2aS$
$\Rightarrow V_B^2=0+2\times 10\times 100$
$\Rightarrow V_B=20\sqrt{5}m/s$

For BC, $V_c=O,V_B=20\sqrt{5}m/s,S=2m$
$S=\left(\frac{v+u}{2}\right)t$ (assuming constant deceleration)
$\Rightarrow t=\frac{2\times 2}{20\sqrt{5}}=\frac{\sqrt{5}}{25}s$