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JEE Main 201910 AprilMorning Shift
Exam
JEE Main
Year
2019
Date
10 April
Shift
Morning Shift
Subject
Physics
Chapter
Laws of Motion

JEE Main 2019 · 10 April · Morning Shift · Physics · Laws of Motion

A ball is thrown upward with an initial velocity V _ 0 from the...

A ball is thrown upward with an initial velocity V0{V}_{0} from the surface of the earth. The motion of the ball is affected by a drag force equal to mγv2m\gamma{v}^{2} (where mm is mass of the ball, vv is its instantaneous velocity and γ\gamma is a constant). Time taken by the ball to rise to its zenith is:

Options

  1. A1\gammagln⁡(1+γgV0)\dfrac{1}{\sqrt{\gammag}}\ln \left(1 + \sqrt{\dfrac{\gamma}{g}}{V}_{0}\right)
  2. B1\gammagtan⁡−1(γgV0)\dfrac{1}{\sqrt{\gammag}}\tan^{- 1} \left(\sqrt{\dfrac{\gamma}{g}}{V}_{0}\right)
  3. C1\gammagsin⁡−1(γgV0)\dfrac{1}{\sqrt{\gammag}}\sin^{- 1} \left(\sqrt{\dfrac{\gamma}{g}}{V}_{0}\right)
  4. D12\gammagtan⁡−1(2γgV0)\dfrac{1}{\sqrt{2\gammag}}\tan^{- 1} \left(\sqrt{\dfrac{2\gamma}{g}}{V}_{0}\right)

Correct answer

B. 1\gammagtan⁡−1(γgV0)\dfrac{1}{\sqrt{\gammag}}\tan^{- 1} \left(\sqrt{\dfrac{\gamma}{g}}{V}_{0}\right)

Step-by-step solution

Net force on ball,
F=−mg−mγv2F = - mg - m\gamma{v}^{2}
a=−g−γv2a = - g - \gamma{v}^{2}
dvdt=−g−γv2\dfrac{dv}{dt} = - g - \gamma{v}^{2}
∫v=V0v=0dvg+γv2=∫t=0t−dt\mathop{∫}_{v = {V}_{0}}^{v = 0}\dfrac{dv}{g + \gamma{v}^{2}} =\mathop{∫}_{t = 0}^{t}- dt
(1gγtan⁡−1γgv)v=V0v=0=−t{\left(\dfrac{1}{\sqrt{g\gamma}}\tan^{- 1}\sqrt{\dfrac{\gamma}{g}}v\right)}_{v = {V}_{0}}^{v = 0} = - t
∴t=1\gammagtan⁡−1(γgV0)∴ t = \dfrac{1}{\sqrt{\gammag}}\tan^{- 1}\left(\sqrt{\dfrac{\gamma}{g}}{V}_{0}\right)

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