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VMMC VMMC Medical Solved Paper-2012

  • question_answer
    The speed of a projectile at its maximum height is half of its initial speed. The angle of projection is

    A) \[{{60}^{o}}\]             

    B)  \[{{15}^{o}}\]

    C) \[{{30}^{o}}\]              

    D) \[{{45}^{o}}\]

    Correct Answer: A

    Solution :

     The speed of a projectile at its maximm. height                                \[v={{v}_{0}}\cos \theta \] \[\frac{{{v}_{0}}}{2}={{v}_{0}}\cos \theta \] \[\cos \theta =\frac{1}{2}\] \[\theta ={{60}^{o}}\]


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