Q 11-05-175JEE MainJEE Main 2020 (4 Sep, Shift 2)Medium
A small ball of mass $m$ is thrown upward with velocity $u$ from the ground. The ball experiences a resistive force $mkv^{2}$ where $v$ is its speed. The maximum height attained by the ball is:
Answer: (D) $\dfrac{1}{2k}\ln\left(1 + \dfrac{ku^{2}}{g}\right)$
On the way up: $v\dfrac{dv}{dh} = -(g + kv^{2})$, so
$$H = \int_0^{u}\frac{v\,dv}{g + kv^{2}} = \frac{1}{2k}\ln\left(\frac{g + ku^{2}}{g}\right) = \frac{1}{2k}\ln\left(1 + \frac{ku^{2}}{g}\right)$$
Solution by Sreeraj P, M.Sc Physics