Question

A body is projected up with a velocity equal to $\left(3 / 4\right)^{th}$ of the escape velocity from the surface of the earth. The height it reaches is, (Radius of earth $=R$ )

• A. $\frac{10 R}{9}$
• B. $\frac{9}{7}R$
• C. $\frac{9}{8}R$
• D. $\frac{10 R}{3}$

Answer 1

Answer: (B)

Escape velocity for the earth is
$v_{e}=\sqrt{\frac{2 G M}{R}}$
Given the velocity projection of the body
$=v=\frac{3}{4}v_{e}=\frac{3}{4} \, \sqrt{\frac{2 G M}{R}}$
Total energy on the earth = Total energy at maximum height $h$
$\Rightarrow \frac{1}{2}mv^{2}+\left(- \frac{G M m}{R}\right)=0+\left(- \frac{G M m}{R + h}\right)$
$\Rightarrow \frac{1}{2}m \, . \, \frac{9}{16}. \, \frac{2 G M}{R}-\frac{G M m}{R}=-\frac{G M m}{R + h}$
$\Rightarrow \frac{9}{16}-1=- \, \frac{R}{R + h}$ or $-\frac{R}{R + h}=\frac{- 7}{16}$
$\Rightarrow 7R+7h=16 \, R$
$7h=9R\Rightarrow h=\frac{9}{7}R$