Question

Two balls $A$ and $B$ are thrown vertically upwards from the same location on the surface of the earth with velocities $2\sqrt{\frac{gR}{3}}$ and $\sqrt{\frac{2gR}{3}}$ 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
$\frac{1}{2}m{v}^2-\frac{GMm}{R}=-\frac{GMm}{(R+h)}$
or $v^2=\frac{GMh}{R(R+h)}=\frac{2ghR}{(R+h)}$
$\left(\because g=\frac{GM}{R^2}\right)$
For ball $A,v=2\sqrt{\frac{gR}{3}},h=h_A$
$\therefore \frac{4gR}{3}=\frac{2g{h}_{A}R}{\left(R+h_A\right)}$
or $h_A=2R$
For ball $B,v=\sqrt{\frac{2gR}{3}},h=h_B$
$\therefore \frac{2gR}{3}=\frac{2g{h}_{B}R}{\left(R+h_B\right)}$
or $h_B=\frac{R}{2}$
$\therefore \frac{h_A}{h_R}=4$