Q 11-02-123JEE MainJEE Main 2019 (10 Apr, Shift 1)Hard
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 $m\gamma v^2$ (where $m$ is mass of the ball, $v$ is its instantaneous velocity and $\gamma$ is a constant). Time taken by the ball to rise to its zenith is:
Answer: (B) $\dfrac{1}{\sqrt{\gamma g}}\tan^{-1}\left(\sqrt{\dfrac\gamma g}\,V_0\right)$
On the way up both gravity and drag act downward:
$$\frac{dv}{dt} = -(g + \gamma v^2)$$
$$t = \int_0^{V_0}\frac{dv}{g+\gamma v^2} = \frac1\gamma\cdot\sqrt{\frac\gamma g}\tan^{-1}\left(\sqrt{\frac\gamma g}\,V_0\right) = \frac{1}{\sqrt{\gamma g}}\tan^{-1}\left(\sqrt{\frac\gamma g}\,V_0\right)$$
Solution by Sreeraj P, M.Sc Physics