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 an reaching the ground. Then \[v_{g}^{2}-{{v}^{2}}=2a{{s}_{2}}\] or \[v_{g}^{2}-{{(1000)}^{2}}=2\times (-2.5)\times 1995\] \[{{v}_{g}}=5m/s\]
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