Question

A ball is thrown upwards 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 constant). Time taken by the ball to rise to its maximum height is

• A. $\frac{1}{\sqrt{2\gamma g}}ta{n}^{-1}\left(\sqrt{\frac{2\gamma }{g}}{v}_0\right)$
• B. $\frac{1}{\sqrt{\gamma g}}ta{n}^{-1}\left(\sqrt{\frac{\gamma }{g}}{v}_0\right)$
• C. $\frac{1}{\sqrt{\gamma g}}si{n}^{-1}\left(\sqrt{\frac{\gamma }{g}}{v}_0\right)$
• D. $\frac{1}{\sqrt{\gamma g}}In(1+\sqrt{\frac{\gamma }{g}}{v}_0)$

Answer 1

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

Given, drag force, F =$m\gamma v^2$ ...(i)
As we know, general equation of force
= ma ....(ii)
Comparing Eqs. (i) and (ii), we get
a = $\gamma v^2$
$\therefore$ Net retardation of the ball when thrown vertically upwards is
$a_{net}=-(g+\gamma v^2)=\frac{dv}{dt}$
$\Rightarrow \frac{dv}{(g+\gamma v^2)}=-dt$
By integrating both sides in known limits, i.e, when the ball thrown upwards with velocity $v_o$ and then reaches to its maximum height, i.e. for maximum height at time = t, v = 0
$\Rightarrow {\int }_{vo}^o\frac{dv}{(\gamma v^2+g)}={\int }_0^t-dt$
or $\frac{1}{\gamma }{\int }_{vo}^0\frac{1}{{\left[\right(\sqrt{\frac{g}{\gamma }})}^2+v^2]}dv=-{\int }_0^{t}dt$
$\Rightarrow \frac{1}{\gamma }.\frac{1}{\sqrt{g/\gamma }}{\left[ta{n}^{-1}\left(\frac{v}{\sqrt{g/\gamma }}\right)\right]}_{v0}^0=-t$
$[\because \int \frac{1}{x^2+a^2}dx=\frac{1}{a}ta{n}^{-1}\left(\frac{x}{a}\right)]$
$\Rightarrow \frac{1}{\sqrt{\gamma g}}.ta{n}^{-1}\left(\frac{\sqrt{\gamma }{v}_0}{\sqrt{g}}\right)=+t$