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 $m\gamma {\upsilon }^2$ (where $m$ is mass of the ball, $\upsilon$ 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. $\frac{1}{\sqrt{\gamma g}}In\left(1+\sqrt{\frac{\gamma }{g}}{V}_0\right)$

Answer 1

Answer: (B)

Given, drag force, $F=my{v}^2\dots ..(i)$
Using Newton's second law,
$F=$ ma.....(ii)
Comparing Eqs. (i) and (ii), we get
$<br/>a=v{v}^2<br/>$
$\therefore$ Net retardation of the ball when thrown vertically upward is
$<br/>\begin{array}{l}<br/>a_{\text{net }}=-\left(g+v{v}^2\right)=\frac{dv}{dt}\\ <br/>\Rightarrow \frac{dv}{\left(g+v{v}^2\right)}=-dt\dots ..\text{ (iii) }<br/>\end{array}<br/>$
By integrating both sides of Eq. (iii) in known limits, i.e. when the ball thrown upward with velocity $v_0$ and then reaches to its zenith i.e for maximum height at time $t=t,v=0$
$\Rightarrow {\int }_{v_0}^0\frac{dv}{\left(v{v}^2+g\right)}={\int }_0^t-dt$
or $\frac{1}{v}{\int }_{v_0}^0\frac{1}{\left[{\left(\sqrt{\frac{g}{Y}}\right)}^2+v^2\right]}dv=-{\int }_0^{t}dt$
$\Rightarrow \frac{1}{V}\cdot \frac{1}{\sqrt{\frac{g}{V}}}\cdot {\left[{\tan }^{-1}\left(\frac{V}{\sqrt{\frac{g}{V}}}\right)\right]}_{v_0}^0=-t$
$\left[\because \int \frac{1}{x^2+a^2}dx=\frac{1}{a}{\tan }^{-1}\left(\frac{x}{a}\right)\right]$
$\Rightarrow -\frac{1}{\sqrt{\gamma g}}\cdot {\tan }^{-1}\left(\frac{\sqrt{\bar{\gamma }}{v}_0}{\sqrt{g}}\right)=-t$
$\Rightarrow t=\frac{1}{\sqrt{yg}}{\tan }^{-1}\left(\frac{\sqrt{\bar{y}}{v}_0}{\sqrt{g}}\right)$