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A ball is thrown upward with an initial velocity V0 from the surface of the earth. The motion of the ball is affected by a drag force equal to mγ 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:

1√(γ g) ^-1⁡ (√((γ )/(g))V0 )

1√(γ g) ⁡ (1+√((γ )/(g))V0 )

1√(γ g) ^-1⁡ (√((γ )/(g))V0 )

1√(2γ g) ^-1⁡ (√((2γ )/(g))V0 )

Laws of Motion

JEE Main level, single correct, medium

43 of 210

Question 43 of 210

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:

Show answer and solution

Answer: (B) $\dfrac{1}{\sqrt{\gamma g}}{\tan }^{-1}⁡\left(\sqrt{\dfrac{\gamma }{g}}{V}_{0}\right)$

Net force on ball,

F=-mg-m\gamma {v}^{2}

a=-g-\gamma {v}^{2}

\dfrac{dv}{dt}=-g-\gamma {v}^{2}

\int _{v={V}_{0}}^{v=0}\dfrac{dv}{g+\gamma {v}^{2}}=\int _{t=0}^{t}-dt

{\left(\dfrac{1}{\sqrt{g\gamma }}{tan}^{-1}\sqrt{\dfrac{\gamma }{g}}v\right)}_{v={V}_{0}}^{v=0}=-t

∴t=\dfrac{1}{\sqrt{\gamma g}}{tan}^{-1}\left(\sqrt{\dfrac{\gamma }{g}}{V}_{0}\right)

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JEE Main 2025 Apr 2 Shift 1 question paper

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