Q 11-07-011NEETNEET 2021Top questionMedium
A particle of mass 'm' is projected with a velocity $v = kV_e$ ($k < 1$) from the surface of the earth. ($V_e$ = escape velocity) The maximum height above the surface reached by the particle is :
Answer: (A) $\dfrac{Rk^2}{1 - k^2}$
Concept. Energy conservation with $V_e^2 = \dfrac{2GM}{R}$.
Step 1. Equation.
$$\frac12mk^2V_e^2 - \frac{GMm}{R} = -\frac{GMm}{R + h} \;\Rightarrow\; k^2\frac{GM}{R} = \frac{GMh}{R(R + h)}$$
Step 2. Solve. $k^2(R + h) = h \Rightarrow h = \dfrac{Rk^2}{1 - k^2}$.
Answer. $\dfrac{Rk^2}{1 - k^2}$ (Option A).
Solution by Sreeraj P, M.Sc Physics