Question

Two balls $A$ and $B$ are thrown vertically upwards from the same location on the surface of the earth with velocities $2 \sqrt{\frac{g R}{3}}$ and $\sqrt{\frac{2 g R}{3}}$ respectively, where $R$ is the radius of the earth and $g$ is the acceleration due to gravity on the surface of the earth. The ratio of the maximum height attained by $A$ to that attained by $B$ is

• A. 2
• B. 4
• C. 8
• D. $4 \sqrt{2}$

Answer 1

Answer: (B)

If $h$ is the maximum height attained, then by the law of conservation of energy
$\frac{1}{2} m v^{2}-\frac{G M m}{R}=-\frac{G M m}{(R+h)}$
or $v^{2}=\frac{G M h}{R(R+h)}=\frac{2 g h R}{(R+h)}$
$\left(\because g=\frac{G M}{R^{2}}\right)$
For ball $A, v=2 \sqrt{\frac{g R}{3}}, h=h_{A}$
$\therefore \frac{4 g R}{3}=\frac{2 g h_{A} R}{\left(R+h_{A}\right)}$
or $h_{A}=2 R$
For ball $B, v=\sqrt{\frac{2 g R}{3}}, h=h_{B}$
$\therefore \frac{2 g R}{3}=\frac{2 g h_{B} R}{\left(R+h_{B}\right)}$
or $h_{B}=\frac{R}{2}$
$ \therefore \frac{h_{A}}{h_{R}}=4$