• # question_answer A ball is thrown upward with an initial velocity ${{V}_{0}}$from the surface of the earth. The motion of the ball is affected by a drag force equal to $m\gamma {{\upsilon }^{2}}$ (where m is mass of the ball, $\upsilon$is its instantaneous velocity and $\gamma$ is a constant). Time taken by the ball to rise to its zenith is : [JEE Main 10-4-2019 Morning] A) $\frac{1}{\sqrt{\gamma g}}{{\sin }^{-1}}\left( \sqrt{\frac{\gamma }{g}}{{V}_{0}} \right)$    B) $\frac{1}{\sqrt{\gamma g}}{{\tan }^{-1}}\left( \sqrt{\frac{\gamma }{g}}{{V}_{0}} \right)$C) $\frac{1}{\sqrt{2\gamma g}}{{\tan }^{-1}}\left( \sqrt{\frac{2\gamma }{g}}{{V}_{0}} \right)$D) $\frac{1}{\sqrt{\gamma g}}\ln \left( 1+\sqrt{\frac{\gamma }{g}}{{V}_{0}} \right)$

$-(g+\gamma {{\text{v}}^{2}})=\frac{d\text{v}}{dt}$           $-gdt=\frac{g}{\gamma }\left( \frac{d\text{v}}{\frac{g}{\gamma }+{{\text{v}}^{2}}} \right)$ Integrating $0\to t\And {{V}_{0}}\to 0$:- $-gt=-\sqrt{\frac{g}{\gamma }}{{\tan }^{-1}}\left( \frac{{{V}_{0}}}{\sqrt{\frac{g}{\gamma }}} \right)$ $t=\frac{1}{\sqrt{\gamma g}}{{\tan }^{-1}}\left( \sqrt{\frac{\gamma }{g}}{{V}_{0}} \right)$