Question

A ball is thrown upward with an initial velocity $v_o$ 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 (maximum height) is

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

Answer 1

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

Given, drag force , $F=m\gamma v^2$.....(i)
Using Newton's second law,
$F=ma$.....(ii)
Comparing Eqs. (i) and (ii) , we get
$a=\gamma v^2$
$\therefore$ Net retardation of the ball when thrown vertically upward is
$a_{net}=-(g+\gamma v^2)=\frac{dv}{dt}$
$\Rightarrow \frac{dv}{(g+\gamma v^2)}=-dt$ .....(iii)
By integrating both sides of Eq. (iii) in known limits, i.e. when the ball thrown upward with velocity $v_o$ and then reaches to its zenith i.e for maximum height at time $t$ = $t$, $v=0$
$\Rightarrow {\int }_{v_o}^0\frac{dv}{(\gamma v^2+g)}={\int }_0^t-dt$
or $\frac{1}{\gamma }{\int }_{v_o}^0\frac{1}{[{\left(\sqrt{\frac{g}{\gamma }}\right)}^2+v^2]}dv=-{\int }_0^{t}dt$
$\Rightarrow \frac{1}{\gamma }.\frac{1}{\sqrt{\frac{g}{\gamma }}}.{\left[{\tan }^{-1}\left(\frac{v}{\sqrt{\frac{g}{\gamma }}}\right)\right]}_{v_o}^0=-t$
$[\because \int \frac{1}{x^2+a^2}dx=\frac{1}{a}{\tan }^{-1}\left(\frac{x}{a}\right)]$
$\Rightarrow -\frac{1}{\sqrt{\gamma g}}.{\tan }^{-1}\left(\frac{\sqrt{\gamma }{v}_o}{\sqrt{g}}\right)=-t$
$\Rightarrow t=\frac{1}{\sqrt{\gamma g}}.{\tan }^{-1}\left(\frac{\sqrt{\gamma }{v}_o}{\sqrt{g}}\right)$