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 m γ u2 (where m is mass of the ball, u is its instantaneous velocity and γ is a constant). Time taken by the ball to rise to its zenith is: (a) (1)/(√(γ g))tan ⁻¹(√((γ)/(g))V₀) (b) (1)/(√(γ g))ln (1 + √((γ)/(g))V₀) (c) (1)/(√(γ g))sin ⁻¹(√((γ)/(g))V₀) (d) (1)/(√(2γ g))tan ⁻¹(√((2γ)/(g))V₀)

Answer: (a). While rising: du/dt = -(g + γ u²). t = ∫₀V₀ (du)/(g + γ u²) = (1)/(√(γ g))tan ⁻¹(√((γ)/(g)) u)|₀V₀ t = (1)/(√(γ g))tan ⁻¹(√((γ)/(g))V₀) So the answer is (a).

A ball is thrown upward with an initial velocity V0 from the surface… | JEE Main 2019 (10 Apr, Shift 1)

Q32JEE Main 2019 · 10 APR · SHIFT 1Difficult

Inertia, First & Second Laws

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 m $\gamma$ u2 (where m is mass of the ball, u is its instantaneous velocity and $\gamma$ is a constant). Time taken by the ball to rise to its zenith is:

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Answer: (a)

While rising: $\frac{du}{dt} = -(g + \gamma u^2)$ .
$t = \int_0^{V_0} \frac{du}{g + \gamma u^2} = \frac{1}{\sqrt{\gamma g}}\tan^{-1}\left(\sqrt{\frac{\gamma}{g}}\,u\right)\Big|_0^{V_0}$
$t = \frac{1}{\sqrt{\gamma g}}\tan^{-1}\left(\sqrt{\frac{\gamma}{g}}V_0\right)$
So the answer is (a).

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Inertia, First & Second Laws question from JEE Main 2019 (10 Apr, Shift 1)

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