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

(B) $\frac{1}{\sqrt{\gamma \mathrm{~g}}} \tan ^{-1}\left(\sqrt{\frac{\gamma}{\mathrm{g}}} \mathrm{V}_{0}\right)$

(A) $\frac{1}{\sqrt{\gamma \mathrm{~g}}} \ln \left(1+\sqrt{\frac{\gamma}{\mathrm{g}}} \mathrm{V}_{0}\right)$

(C) $\frac{1}{\sqrt{\gamma g}} \sin ^{-1}\left(\sqrt{\frac{\gamma}{g}} \mathrm{V}_{0}\right)$

(D) $\frac{1}{\sqrt{2 \gamma \mathrm{~g}}} \tan ^{-1}\left(\sqrt{\frac{2 \gamma}{\mathrm{~g}}} \mathrm{v}_{0}\right)$

JEE Main · 2019 · 10 Apr 2019 · Shift 1 · Laws of Motion

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 Instantneous velocity and

γ

is a constant). Time taken by the ball to rise to its zenith is:

Answer: (B)

Taking the upward flight with quadratic drag, the equation of motion is

- m (g + γ v^{2}) = m \frac{d v}{d t}

, so

\frac{d v}{g + γ v ^{2}} = - d t

. Writing it as

\frac{d v}{\frac{g}{γ} + v ^{2}} = - γ d t

and integrating the speed from

V_{0}

down to

0

while time runs from

0

t

, the left side gives an arctangent,

\frac{1}{g / γ} tan^{- 1} (\frac{v}{g / γ})

between the limits. This leaves

t = \frac{1}{γ g} tan^{- 1} (\frac{γ}{g} V_{0})

JEE Main · 2019 · 10 Apr 2019 · Shift 1 · Single correct option