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BHU PMT BHU PMT (Mains) Solved Paper-2009

  • question_answer
    A parachutist drops freely from an airplane for 10 s before the parachute opens. He then descends with a uniform retardation of 2.5 \[m{{s}^{-2}}\]. If he bails out of the plane at a height of 2495 m and g is\[10\text{ }m{{s}^{-2}},\] his velocity on reaching the ground will be

    A)  \[5m{{s}^{-1}}\]                              

    B)  \[10\,m{{s}^{-1}}\]

    C)  \[15m{{s}^{-1}}\]                           

    D)  \[20\,m{{s}^{-1}}\]

    Correct Answer: A

    Solution :

                     Initial velocity of dropping = zero                 Let \[{{v}_{1}}\]be velocity at end of 10s                 \[\Rightarrow \]               \[{{v}_{1}}=gt\]                 \[\Rightarrow \]               \[{{v}_{1}}=100\,\,m{{s}^{-1}}\]                 Distance travelled during this time is                 \[{{h}_{1}}=\frac{v_{1}^{2}}{2g}=\frac{{{(100)}^{2}}}{2(10)}\]                 \[\Rightarrow \]               \[{{h}_{1}}=500\,m\] So, a remaining distance of\[2495-500=1995\]m has to be travelled with a retardation of\[2.5\,m{{s}^{-2}}\]. Let the parachutist strike the ground with velocity v.                 Then   \[{{v}^{2}}-v{{'}^{2}}=2a(h-h')\]                 \[\Rightarrow \]\[{{v}^{2}}-{{(100)}^{2}}=2(-2.5)(1995)\]                 \[\Rightarrow \] \[{{v}^{2}}=10000-9975\]                 \[\Rightarrow \] \[{{v}^{2}}=25\]                 \[\Rightarrow \]               \[v=5\,m{{s}^{-1}}\]


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