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JIPMER Jipmer Medical Solved Paper-2015

  • question_answer
    A body dropped from a height h with initial velocity zero, strikes the ground with a velocity 3 m/s. Another body of same mass dropped from the same height h with an initial velocity of 4 m/s. Find the final velocity of second mass, with which it strikes the group.

    A)  3 m/s                   

    B)         4 m/s

    C)  5 m/s                   

    D)         12 m/s

    Correct Answer: C

    Solution :

    Initial velocity of first body \[({{u}_{1}})=0;\] Final velocity \[({{v}_{1}})\] = 3 m/s and initial velocity of second body \[({{u}_{2}})\] = 4 m/s. We know that height \[(h)=\frac{v_{1}^{2}}{2g}=\frac{{{(3)}^{2}}}{2\times 9.8}=0.46\,m.\] Therefore velocity of the second body, \[{{y}_{2}}=\sqrt{u_{2}^{2}+2gh}=\sqrt{{{(4)}^{2}}+2\times 9.8\times 0.46}\]      \[=5\text{ }m\text{/}s.\]


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