Question

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

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

Answer 1

The correct option is B $\frac{9}{7}R$

Escape velocity for the earth is,

$v_s=\sqrt{\frac{2GM}{R}}$

Given, the velocity projection of the body

$v=\frac{3}{4}{v}_s=\frac{3}{4}\sqrt{\frac{2GM}{R}}$

Total energy on the earth's surface
= Total energy at maximum height $h$

$\Rightarrow \frac{1}{2}m{v}^2+(-\frac{GMm}{R})=0+(-\frac{GMm}{R+h})$

$\Rightarrow \frac{1}{2}\times m\times \frac{9}{16}\frac{2GM}{R}-\frac{GMm}{R}=-\frac{GMm}{R+h}$

$\Rightarrow (\frac{9}{16}-1)=-\frac{R}{R+h}$

$\Rightarrow -\frac{R}{R+h}=\frac{-7}{16}$

$\Rightarrow 7R+7h=16R$

$\Rightarrow h=\frac{9}{7}R$

Hence, option $\left(B\right)$ is correct.