A body falls freely towards the earth from a height 2R, above the surface of the earth, where initially it was at rest. If R is the radius of the earth then its velocity on reaching the surface of the earth is

$\sqrt{\frac{4}{3} g R}$

Initial energy of the body = $-\frac{GMm}{(R + 2 R)}$

Final energy of the body = $-\frac{GMm}{R}+\frac{1}{2} mv^2$

where m is the mass of the body and M is the mass of the earth.

Applying conservation of mechanical energy

$-\frac{\mathrm{GMm}}{3 \mathrm{R}}=-\frac{\mathrm{GMm}}{\mathrm{R}}+\frac{1}{2} \mathrm{mv}^2$

$\Rightarrow \frac{1}{2} \mathrm{mv}^2=\frac{2 \mathrm{GMm}}{3 \mathrm{R}}$

$\Rightarrow \mathrm{v}^2=\frac{4 \mathrm{GM}}{3 \mathrm{R}} \Rightarrow v = \sqrt{\frac{4GM}{3R}}$

$=\sqrt{\frac{4 \mathrm{GMR}}{3 \mathrm{R}^2}}=\sqrt{\frac{4}{3} \mathrm{gR}}$