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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 $m γ v^{2}$ (where m is mass of the ball, v is its instantaneous velocity and $γ$ is a constant). Time taken by the ball to rise to its zenith is :

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$\frac{1}{\sqrt{γ g}} \tan^{- 1} (\sqrt{\frac{γ}{g}} v_{0})$

$\frac{1}{\sqrt{γ g}} \sin^{- 1} (\sqrt{\frac{γ}{g}} v_{0})$

$\frac{1}{\sqrt{γ g}} \ln (1 + \sqrt{\frac{γ}{g}} v_{0})$

$\frac{1}{\sqrt{2 γ g}} \tan^{- 1} (\sqrt{\frac{2 γ}{g}} v_{0})$

answer is A.

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Detailed Solution

Net acceleration $\frac{d v}{d t} = a = - (g + γ v^{2})$

Let time t required to rise to its zenith (v = 0) so,

$\int_{v_{0}}^{0} \frac{- d v}{g + γ v^{2}} = \int_{0}^{t} d t [for H_{\max,} v = 0]$
$∴ t = \frac{1}{\sqrt{γ g}} \tan^{-} (\frac{\sqrt{γ}}{\sqrt{g}} v_{0})$

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