A projectile is fired from the surface of earth of radius $R$ with a velocity $k v_{e}$ $($ where $v_{e}$ , is the escape velocity from surface of earth and $k < 1)$ . Neglecting air resistance, the maximum height of rise from the centre of earth is

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

$\frac{R}{k^{2} - 1}$

$k^{2} R$

$\frac{R}{1 - k^{2}}$

$k R$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$K_{i} + U_{i} = K_{f} + U_{f}$

$\frac{1}{2} m {(k v_{e})}^{2} - \frac{G M m}{R} = 0 - \frac{G M m}{r}$

$∴ \frac{1}{2} k^{2} (2 g R) - \frac{G M}{R} = - \frac{G M}{r}$

or $\frac{G M}{r} (1 - k^{2}) = \frac{G M}{r} (as g R = \frac{G M}{R})$

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

JEE

NEET

Foundation JEE

Foundation NEET

CBSE