Question

A projectile is fired upward from the surface of the earth with a velocity $k{v}_e$, where $v_e$ is escape velocity and $k<1$. If $R$ is the radius of earth, then neglecting the air resistance, the maximum height to which it will rise is

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

Answer 1

The correct option is D $\frac{k^{2}R}{1-k^2}$

If a body is projected from the surface of earth with a velocity $v$ and reaches a height $h$, then by conservation of energy (relative to surface of earth)
$\frac{1}{2}m{v}^2=\frac{mgh}{[1+\left(\frac{h}{R}\right)]}$
In this case, $v=k{v}_e=k\sqrt{2gR}$
So, $\frac{1}{2}{k}^2\left(2gR\right)=\frac{gh}{[1+\left(\frac{h}{R}\right)]}or{k}^2=\frac{h}{R+h}$
$or{k}^{2}R+k^{2}h=h\Rightarrow h=\frac{k^{2}R}{1-k^2}$