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JEE Main & Advanced Physics Question Bank Oblique Projectile Motion

  • question_answer
    For a projectile, the ratio of maximum height reached to the square of flight time is (g = 10 ms?2)  [EAMCET (Med.) 2000]

    A)             5 : 4

    B)             5 : 2

    C)             5 : 1    

    D)             10 : 1

    Correct Answer: A

    Solution :

                    \[H=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}\] and \[T=\frac{2u\sin \theta }{g}\] So \[\frac{H}{{{T}^{2}}}=\frac{{{u}^{2}}{{\sin }^{2}}\theta /2g}{4{{u}^{2}}{{\sin }^{2}}\theta /{{g}^{2}}}=\frac{g}{8}=\frac{5}{4}\]  


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