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JEE Main & Advanced Physics Question Bank Oblique Projectile Motion

  • question_answer
    A projectile thrown with a speed \[v\] at an angle \[\theta \] has a range \[R\] on the surface of earth. For same \[v\] and \[\theta \], its range on the surface of moon will be

    A)             \[R/6\] 

    B)             \[6R\]

    C)             \[R/36\]           

    D)             \[36R\]

    Correct Answer: B

    Solution :

                    Range is given by \[R=\frac{{{u}^{2}}\sin 2\theta }{g}\] On moon \[{{g}_{m}}=\frac{g}{6}\]. Hence \[{{R}_{m}}=6R\]


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