Question

A projectile is projected with velocity $k{v}_e$ in vertically upward direction from the ground into the space $v_e$ is escape velocity and (k < 1). If air resistance is considered to be negligible then the maximum height from the centre of earth to which it can go will be (R = radius of earth)

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

From the law of conservation of energy Difference in potential energy between ground and maximum height = Kinetic energy at the point of projection

$\frac{mgh}{1+\frac{h}{R}}=\frac{1}{2}m(k{v}_e{)}^2=\frac{1}{2}m{k}^2{v}_e^2=\frac{1}{2}m{k}^2(\sqrt{2gR}{)}^2$

[As $v_e=\sqrt{2gR}$]

By solving height from the surface of earth, h = $\frac{R{k}^2}{1-k^2}$

So height from the center of earth, r = R + h = R + $\frac{R{k}^2}{1-k^2}=\frac{R}{1-k^2}$.