Question

A projectile rises upto a maximum distance of $\frac{R}{1-k^2}$ where K is a constant and R is the radius of Earth. If the velocity of the projectile with which it should be fired upwards from the surface of Earth to reach this height is equal to the product of a coefficient and escape velocity, then this coefficient is equal to:

• A. K
• B. $K^2$
• C. $K^{\frac{3}{2}}$
• D. $K^{\frac{1}{2}}$

Answer 1

The correct option is A

K

Let the velocity with which the projectile is fired be 'v'.

Here ,$r=\frac{R}{1-k^2}$.

Conserving mechanical energy, initially & at its highest point,

$\frac{1}{2}m{v}^2-\frac{GmM}{R}=\frac{-GmM}{r}$,

Where, m = mass of the projectile,

M = mass of the Earth.

$\frac{1}{2}{v}^2=\frac{GM}{R}-\frac{GM(1-k^2)}{R}$

$v^2=\frac{2GM}{R}\left\{k^2\right\}\Rightarrow v=\sqrt{\frac{2GM}{R}}\left(k\right)$.

we already know,

$v_e=\frac{2GM}{R}$

$\Rightarrow v=k{v}_e$.

Hence (a) correct option