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, $F=m y v^2 \ldots . .(i)$
Using Newton's second law,
$F=$ ma.....(ii)
Comparing Eqs. (i) and (ii), we get
$
a=v v^2
$
$\therefore$ Net retardation of the ball when thrown vertically upward is
$
\begin{array}{l}
a_{\text {net }}=-\left(g+v v^2\right)=\frac{d v}{d t} \\
\Rightarrow \frac{d v}{\left(g+v v^2\right)}=-d t \ldots . . \text { (iii) }
\end{array}
$
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 $t=t, v=0$
$\Rightarrow \int_{v_0}^0 \frac{d v}{\left(v v^2+g\right)}=\int_0^t-d t$
or $\frac{1}{ v } \int_{v_0}^0 \frac{1}{\left[\left(\sqrt{\frac{ g }{ Y }}\right)^2+ v ^2\right]} d v=-\int_0^{ t } d t$
$\Rightarrow \frac{1}{ V } \cdot \frac{1}{\sqrt{\frac{ g }{ V }}} \cdot\left[\tan ^{-1}\left(\frac{ V }{\sqrt{\frac{ g }{ V }}}\right)\right]_{ v _0}^0=- t$
$\left[\because \int \frac{1}{x^2+a^2} d x=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)\right]$
$\Rightarrow-\frac{1}{\sqrt{ \gamma g }} \cdot \tan ^{-1}\left(\frac{\sqrt{\bar{\gamma}} v_0}{\sqrt{g}}\right)=- t$
$\Rightarrow t =\frac{1}{\sqrt{ yg }} \tan ^{-1}\left(\frac{\sqrt{\bar{y}} v _0}{\sqrt{ g }}\right)$