Question

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

• A. $\frac{10R}{9}$
• B. $\frac{9R}{7}$
• C. $\frac{9R}{16}$
• D. $\frac{10R}{3}$

Answer 1

Answer: (B)

$\frac{9R}{7}$

By conservation of energy, the total energy at earth's surface is equal to the total potential energy at maximum height $h$.

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

$\left(1/2\right)v^2=GM(\frac{1}{R}-\frac{1}{R+h})=\frac{ghR}{R+h}$ (gravitational

acceleration at earth's surface, $g=GM/R^2$)

Given, $v=\left(3/4\right)v_e=\left(3/4\right)\sqrt{2gR}$

So, $\left(1/2\right)\left(9/16\right)\left(2gR\right)=ghR/R+h$

or $16h=9R+9h$ or $h=9R/7$

So option B will be correct.