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 \[{{v}^{2}}-v{{'}^{2}}=2a(h-h')\] \[\Rightarrow \]\[{{v}^{2}}-{{(100)}^{2}}=2(-2.5)(1995)\] \[\Rightarrow \] \[{{v}^{2}}=10000-9975\] \[\Rightarrow \] \[{{v}^{2}}=25\] \[\Rightarrow \] \[v=5\,m{{s}^{-1}}\]You need to login to perform this action.
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