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JCECE Engineering JCECE Engineering Solved Paper-2011

  • question_answer
    A parachutist, drops first freely from an aero plane for \[10\,\,s\] and then parachute opens out. Now he descends with a net retardation of\[2.5\,\,m/{{s}^{2}}\]. If. he bails out of the plane at a height of \[2495\,\,m\] and \[g=10\,\,m/{{s}^{2}}\], his velocity on reaching the ground will be

    A) \[5\,\,m/s\]                       

    B) \[10\,\,m/s\]

    C)  \[15\,\,m/s\]                   

    D)  \[20\,\,m/s\]

    Correct Answer: A

    Solution :

    The velocity v acquired by the parachutist after\[10\,\,s\]                 \[v=u+gt=0\]                 \[=0+10\times 10=100\,\,m/s\] Then,    \[{{S}_{1}}=ut+\frac{1}{2}g{{t}^{2}}\]                       \[=0+\frac{1}{2}\times 10\times {{10}^{2}}=500\,\,m\] The distance travelled by the parachutist under retardation.                 \[{{S}_{2}}=2495-500=1995\,\,m\] Let \[{{v}_{g}}\] be this velocity a reaching the ground. Then      \[v_{g}^{2}-{{v}^{2}}=2a{{s}_{2}}\] or            \[v_{g}^{2}-{{(100)}^{2}}=2\times (-2.5)\times 1995\]                 \[{{v}_{g}}=5\,\,m/s\]


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