Question

A body is thrown from the surface of the earth with velocity 'u' m/s. The maximum height in m above the surface of the earth upto which it will reach is (R = radius of earth, g = acceleration due to gravity)

• A. $\frac{u^{2}R}{2gR-u^2}$
• B. $\frac{2u^{2}R}{gR-u^2}$
• C. $\frac{u^2{R}^2}{2g{R}^2-u^2}$
• D. $\frac{u^{2}R}{gR-u^2}$

Answer 1

Answer: (A)

$\frac{u^{2}R}{2gR-u^2}$

$g=\frac{G{M}_e}{R^2}$

Applying law of conservation of mechanical energy:-

$\frac{1}{2}m{u}^2-\frac{G{M}_{e}m}{R}=-\frac{G{M}_{e}m}{R+h}$

$⟹\frac{u^2}{2}-gR=\frac{-g{R}^2}{R+h}$

$⟹\frac{g{R}^2}{R+h}=\frac{2gR-u^2}{2}$

$⟹R+h=\frac{2g{R}^2}{2gR-u^2}$

$⟹h=\frac{2g{R}^2}{2gR-u^2}-R$

$⟹h=\frac{u^{2}R}{2gR-u^2}$

Hence, answer is option-(D).