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JEE Main & Advanced Physics Question Bank Oblique Projectile Motion

  • question_answer
    If time of flight of a projectile is 10 seconds. Range is 500 meters. The maximum height attained by it will be                                      [RPMT 1997]

    A)             125 m

    B)               50 m

    C)             100 m

    D)               150 m

    Correct Answer: A

    Solution :

                    \[T=\frac{2u\sin \theta }{g}=10\sec \Rightarrow u\sin \theta =50\ m/s\]             \ \[H=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}=\frac{{{(u\sin \theta )}^{2}}}{2g}=\frac{50\times 50}{2\times 10}=125m\]


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