JEE Main & Advanced Physics Question Bank Oblique Projectile Motion

  • question_answer
    Referring to above question, the angle with the horizontal at which the projectile was projected is [CPMT 1981]

    A)             \[{{\tan }^{-1}}(3/4\])

    B)             \[{{\tan }^{-1}}(4/3)\]

    C)             \[{{\sin }^{-1}}(3/4\])

    D)             Not obtainable from the given data

    Correct Answer: B

    Solution :

                    The angle of projection is given by \[\theta ={{\tan }^{-1}}\left( \frac{{{v}_{y}}}{{{v}_{x}}} \right)={{\tan }^{-1}}\left( \frac{4}{3} \right)\]


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