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 \[=4s+18s=22s\]You need to login to perform this action.
You will be redirected in
3 sec