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AMU Medical AMU Solved Paper-2007

  • question_answer
    A ball is projected horizontally with a velocity of 4 m/s. The velocity of the ball after 0.7 s is \[\left( \text{g}=\text{1}0\text{m}/{{\text{s}}^{\text{2}}} \right)\]

    A)  11 m/s                                

    B)  10 m/s

    C)  8 m/s                                   

    D)  3 m/s

    Correct Answer: C

    Solution :

                                    Velocity of the ball after time t                 \[=\sqrt{v_{x}^{2}+v_{y}^{2}}\]                 \[=\sqrt{{{(4)}^{2}}+{{(gt)}^{2}}}\]                 \[=\sqrt{16+{{(10\times 0.7)}^{2}}}\]                 = 8 m/s


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