Question

A particle of mass $m$ is dropped from a height $2R$ from the surface of the earth. The speed with which the particle hits the surface of the earth is:

• A. $2\sqrt{\frac{gR}{3}}$
• B. $2\sqrt{gR}$
• C. $4\sqrt{\frac{gR}{3}}$
• D. $2\sqrt{\frac{gR}{5}}$

Answer 1

The correct option is A $2\sqrt{\frac{gR}{3}}$

Applying energy conservation principle between the topmost point and the surface of earth, we have:

Loss in potential energy of the system $=$ Increase in kinetic energy of the particle.

$\Delta U=\Delta KE$

$\frac{mgh}{1+\frac{h}{R}}=\frac{1}{2}m{v}^2$

$\Rightarrow \frac{mg\left(2R\right)}{1+\frac{2R}{R}}=\frac{1}{2}m{v}^2$

$v=2\sqrt{\frac{gR}{3}}$

Hence, option $\left(A\right)$ is the correct answer.