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 is taken by the ball to rise to its zenith is:

• A. $\frac{1}{\gamma g}{\tan }^{-1}\left(\sqrt{\frac{\gamma }{g}}{V}_0\right)$
• B. $\frac{1}{\gamma g}{\sin }^{-1}\left(\sqrt{\frac{\gamma }{g}}{V}_0\right)$
• C. $\frac{1}{\gamma g}{\ln }^{-1}(1+\sqrt{\frac{\gamma }{g}}{V}_0)$
• D. $\frac{1}{2\gamma g}{\tan }^{-1}\left(\sqrt{\frac{2\gamma }{g}}{V}_0\right)$

Answer 1

The correct option is A $\frac{1}{\sqrt{\gamma g}}{\tan }^{-1}\left(\sqrt{\frac{\gamma }{g}}{V}_0\right)$

Net upward force acting on the ball,

$F_{net}=-mg-m\gamma v^2$

So, the net acceleration

$\frac{dv}{dt}=a=\frac{F_{net}}{m}=-(g+\gamma v^2)$

Let time $t$ required to rise to its zenith, where $v=0$, so,

${\int }_{V_0}^0\frac{-dv}{g+\gamma v^2}={\int }_0^{t}dt$

Using the integration formula we get,

$-{\left[\frac{1}{\sqrt{\gamma g}}{\tan }^{-1}\left(\frac{\sqrt{\gamma }v}{\sqrt{g}}\right)\right]}_{V_0}^0=t$

$\Rightarrow t=\frac{1}{\sqrt{\gamma g}}{\tan }^{-1}\left(\frac{\sqrt{\gamma }{V}_0}{\sqrt{g}}\right)$

Hence, $\left(A\right)$ is the correct solution