JEE Main & Advanced JEE Main Paper (Held on 10-4-2019 Morning)

  • question_answer 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 \[m\gamma {{\upsilon }^{2}}\] (where m is mass of the ball, \[\upsilon \]is its instantaneous velocity and \[\gamma \] is a constant). Time taken by the ball to rise to its zenith is : [JEE Main 10-4-2019 Morning]

    A) \[\frac{1}{\sqrt{\gamma g}}{{\sin }^{-1}}\left( \sqrt{\frac{\gamma }{g}}{{V}_{0}} \right)\]    

    B) \[\frac{1}{\sqrt{\gamma g}}{{\tan }^{-1}}\left( \sqrt{\frac{\gamma }{g}}{{V}_{0}} \right)\]

    C) \[\frac{1}{\sqrt{2\gamma g}}{{\tan }^{-1}}\left( \sqrt{\frac{2\gamma }{g}}{{V}_{0}} \right)\]

    D) \[\frac{1}{\sqrt{\gamma g}}\ln \left( 1+\sqrt{\frac{\gamma }{g}}{{V}_{0}} \right)\]

    Correct Answer: B

    Solution :

    \[-(g+\gamma {{\text{v}}^{2}})=\frac{d\text{v}}{dt}\]           \[-gdt=\frac{g}{\gamma }\left( \frac{d\text{v}}{\frac{g}{\gamma }+{{\text{v}}^{2}}} \right)\] Integrating \[0\to t\And {{V}_{0}}\to 0\]:- \[-gt=-\sqrt{\frac{g}{\gamma }}{{\tan }^{-1}}\left( \frac{{{V}_{0}}}{\sqrt{\frac{g}{\gamma }}} \right)\] \[t=\frac{1}{\sqrt{\gamma g}}{{\tan }^{-1}}\left( \sqrt{\frac{\gamma }{g}}{{V}_{0}} \right)\]

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