Question

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

• 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. $\frac{1}{\sqrt{\gamma g}}\ln \left(1+\sqrt{\frac{\gamma }{g}}{V}_0\right)$

Answer 1

Answer: (B)

$-\left(g+\gamma v^2\right)=\frac{dv}{dt}$
$-gdt=\frac{g}{\gamma }\left(\frac{dv}{\frac{g}{\gamma }+v^2}\right)$
Intergrating $0\to t\&V_0\to 0$ :-
$-gt=-\sqrt{\frac{g}{\gamma }}{\tan }^{-1}\left(\frac{V_0}{\sqrt{\frac{g}{\gamma }}}\right)$
$t=\frac{1}{\sqrt{\gamma g}}{\tan }^{-1}\left(\sqrt{\frac{\gamma }{g}}{V}_0\right)$