Question

A particle is projected vertically upwards with a velocity $v=\sqrt{gR}$, where $R$ denotes the radius of earth and $g$ is acceleration due to gravity of the surface of earth. The maximum height ascended by the particle is

• A. $R/2$
• B. $R$
• C. $R/4$
• D. $\left(3/4\right)R$

Answer 1

The correct option is B. $R$.

From Conservation of Energy,

$K{E}_i+U_i=K{E}_f+U_f$

$\frac{1}{2}m{v}^2+(-\frac{GMm}{R})=-\frac{GMm}{(R+h)}$

Since, $v=\sqrt{gR}$ and $g=\frac{GM}{R^2}$

$\frac{mgR}{2}-mgR=-mg\frac{R^2}{(R+h)}$

$-\frac{R}{2}=-\frac{R^2}{(R+h)}$

$R+h=2R$

$h=R$.