Question

A ball of mass m is fired vertically upwards from the surface of the earth with velocity $n{v}_e$, where $v_e$ 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

Answer: (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 $v=n{v}_e$

$\frac{1}{2}m{n^2{v}_e}^2=\frac{mghR}{R+h}$

$\frac{1}{2}m{n}^2{\left(\sqrt{2gR}\right)}^2=\frac{mghR}{R+h}$

$\frac{1}{2}m{n}^{2}2gR=\frac{mghR}{R+h}$

$n^2=\frac{h}{R+h}$

$\therefore h=\frac{R{n}^2}{(1-n^2)}$