JEE Main & Advanced Physics Two Dimensional Motion Question Bank Self Evaluation Test - Motion in a Plane

  • question_answer
    For a stone thrown from a lower of unknown height, the maximum range for a projection speed of 10 m/s is obtained for a projection angle of \[30{}^\circ .\] The corresponding distance between the foot of the lower and the point of landing of the stone is

    A) \[10\text{ }m\]              

    B) \[~20\text{ }m\]

    C) \[\left( \text{20/}\sqrt{3} \right)\text{ m  }\!\!~\!\!\text{ }\]

    D)        \[\left( 10/\sqrt{3} \right)\text{ m}\]

    Correct Answer: D

    Solution :

    [d] \[\tan \theta =\frac{{{\text{u}}^{2}}}{\text{Rg}}\] \[\Rightarrow \text{R}\,\,\text{=}\frac{{{\text{u}}^{2}}}{g\tan \theta }=\frac{100}{10\times \sqrt{3}}=\frac{10}{\sqrt{3}}\text{ m}\]


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