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

Answer 1

Answer: (B)

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