Kinematics Ques 9
- 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 taken by the ball to rise to its zenith is
(2019 Main, 10 April I)
(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) $\frac{1}{\sqrt{\gamma g}} \ln \left(1+\sqrt{\frac{\gamma}{g}} v_0\right)$
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Answer:
Correct Answer: 9.(b)
Solution:
Formula:
Instantaneous Acceleration (at an instant):
- Given, drag force, $F=m \psi^{2} \cdots(i)$
As we know, general equation of force
$$ =m a \cdots(ii) $$
Comparing Eqs. (i) and (ii), we get
$$ a=\gamma v^{2} $$
$\therefore$ Net retardation of the ball when thrown vertically upward is
$$ \begin{aligned} a _{net} & =-\left(g+\gamma^{2}\right)=\frac{d v}{d t} \\ \Rightarrow \quad \frac{d v}{\left(g+\gamma^{2}\right)} & =-d t \cdots(iii) \end{aligned} $$
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 $t=t, v=0$
$$ \begin{gathered} \Rightarrow \quad \int _{v _0}^{0} \frac{d v}{\left(Y v^{2}+g\right)}=\int _0^{t}-d t \\ \text { or } \quad \frac{1}{Y} \int _{v _0}^{0} \frac{1}{\sqrt{\frac{g}{Y}}{ }^{2}+v^{2}} d v=-\int _0^{t} d t \\ \Rightarrow \quad \frac{1}{Y} \cdot \frac{1}{\sqrt{g / Y}} \cdot \tan ^{-1} \frac{v}{\sqrt{g / Y}}=-t \\ \Big[\because \int _{v _0}^{x^{2}+a^{2}} d x=\frac{1}{a} \tan ^{-1} \frac{x}{a}\Big] \end{gathered} $$
$$ \Rightarrow \quad \frac{1}{\sqrt{\mathrm{\gamma g}}} \cdot \tan ^{-1} \frac{\sqrt{Y} v _0}{\sqrt{g}}=t $$
BETA
