Question:
A ball is thrown upward with an initial velocity $\mathrm{V}_{0}$ 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:
$\frac{1}{\sqrt{\gamma g}} \tan ^{-1}\left(\sqrt{\frac{\gamma}{g}} \mathrm{~V}_{0}\right)$
$\frac{1}{\sqrt{\gamma} g} \sin ^{-1}\left(\sqrt{\frac{\gamma}{g}} \mathrm{~V}_{0}\right)$
$\frac{1}{\sqrt{\gamma g}} \ln \left(1+\sqrt{\frac{\gamma}{g}} \mathrm{~V}_{0}\right)$
$\frac{1}{\sqrt{2 \gamma} g} \tan ^{-1}\left(\sqrt{\frac{2 \gamma}{g}} \mathrm{~V}_{0}\right)$
Question of from chapter.
JEE Main Previous Year 10 April 2019
Correct Option: 1
Solution:
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