question_answer

After jumping out from the plane, a parachutist falls 80 m without friction. When he opens up the parachute, he decelerates at\[2\text{ }m\text{ }{{s}^{-2}}\]. He reaches the ground with a speed of\[4m\text{ }{{s}^{-1}}\]. How long did the parachutist spend his time in the air? (Take\[\text{g=10 m }{{\text{s}}^{\text{-2}}}\])

A) 4s                                

B) 16 s                 

C) 18 s                 

D)        22 s

Correct Answer: D

Solution :

Time taken when he falls 80 m in the air is \[\therefore \]\[d={{u}_{1}}{{t}_{1}}+\frac{1}{2}{{a}_{1}}t_{1}^{2}\] or \[80=0+\frac{1}{2}\times 10\times t_{1}^{2}\] Or \[t_{1}^{2}=16;{{t}_{1}}=4s\Rightarrow {{v}_{1}}={{u}_{1}}+{{a}_{1}}{{t}_{1}}\] \[{{v}_{1}}=0+10\times 4=40m{{s}^{1}}\] Since \[{{v}_{1}}={{u}_{2}}\therefore {{u}_{2}}=40m{{s}^{-1}}\] Time taken after he opens up his parachute and before reaching the ground, \[{{t}_{2}}\], \[{{v}_{2}}=4m{{s}^{-1}},{{a}_{2}}=-2m{{s}^{-2}}\] \[{{v}_{2}}={{u}_{2}}+{{a}_{2}}{{t}_{2}}\Rightarrow 4=40+(-2)\times {{t}_{2}}\] \[2{{t}_{2}}=36;{{t}_{2}}=18s\] Total time \[=4s+18s=22s\]