A body is dropped from a height of $2$ m. It penetrates into the sand on the ground through a distance of $10$ cm before coming to rest. What is the retardation of the body in sand?

The correct option is B $196{ms}^{-2}$

Initial velocity of the body $u=0$

Velocity of the body just before reaching the sand $V=\sqrt{2gH}$

$⟹V=\sqrt{2\times 9.8\times 2}=\sqrt{4\times 9.8}m/s$

Final velocity in the sand $V_f=0$

Distance covered in sand $S=10cm=0.1m$

Using $V_f^2-V^2=2aS$

Or $0-(\sqrt{4\times 9.8}{)}^2=2\times a\times 0.1$

$⟹a=-196m/s^2$

Retardation is $196m/s^2$