Q 11-07-127JEE MainJEE Main 2022 (26 Jul, Shift 2)Medium
A body is projected vertically upwards from the surface of earth with a velocity equal to one third of escape velocity. The maximum height attained by the body will be (Take radius of earth $= 6400\ \text{km}$ and $g = 10\ \text{m s}^{-2}$)
Answer: (A) $800\ \text{km}$
$v^2 = \dfrac{v_e^2}{9} = \dfrac{2gR}{9}$. Energy conservation:
$$\frac12mv^2 = \frac{GMm}{R} - \frac{GMm}{R+h} = \frac{mgRh}{R+h}\ \Rightarrow\ \frac{gR}{9} = \frac{gRh}{R+h}$$
$$R + h = 9h\ \Rightarrow\ h = \frac R8 = 800\ \text{km}$$
Solution by Sreeraj P, M.Sc Physics