Question

A ball of mass $m$ is fired vertically upwards from the surface of the earth with velocity $n{\nu }_e$, where ${\nu }_e$ is the escape velocity and $n>1$. To what height will the ball rise? Neglecting air resistance, take radius of the earth as $R$

• A. $\frac{R}{n^2}$
• B. $\frac{R}{(1-n^2)}$
• C. $\frac{R{n}^2}{(1-n^2)}$
• D. $R{n}^2$

Answer 1

The correct option is C $\frac{R{n}^2}{(1-n^2)}$

$\frac{1}{2}m{v}^2=\frac{GMm}{R}-\frac{GMm}{R+h}$
$=\frac{GMmh}{R(R+h)}$
$=\frac{mghR}{R+h}$
But, $v_e=\sqrt{2gR}$
and $v=n{v}_e$
$\frac{1}{2}m{n^2{v}_e}^2=\frac{mghR}{R+h}$
$\frac{1}{2}m{n}^2{\left(\sqrt{2gR}\right)}^2=\frac{mghR}{R+h}$
$\frac{1}{2}m{n}^{2}2gR=\frac{mghR}{R+h}$
$n^2=\frac{h}{R+h}$
$\therefore h=\frac{R{n}^2}{(1-n^2)}$