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 \[v_{g}^{2}-{{(100)}^{2}}=2\times (-2.5)\times 1995\] \[{{v}_{g}}=5\,\,m/s\]You need to login to perform this action.
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