A) \[\sqrt{\frac{2.7}{9.8}}s\]
B) \[\sqrt{\frac{5.4}{9.8}}s\]
C) \[\sqrt{\frac{5.4}{8.6}}s\]
D) \[\sqrt{\frac{5.4}{11}}s\]
Correct Answer: D
Solution :
Net acceleration on the elevator car is more than acceleration due to gravity. Since elevator car is ascending upwards, from Newtons second law, the net force is acting upwards hence resultant acceleration is \[{{a}_{r}}=(g+a)\] \[=(9.8+1.2)=11m/{{s}^{2}}\] Relative velocity of observer to elevator is \[{{u}_{r}}=0\] from the equation \[s={{u}_{r}}t+\frac{1}{2}a,{{t}^{2}}\] \[2.7=0+\frac{1}{2}\times 11\times {{t}^{2}}\] \[{{t}^{2}}=\frac{5.4}{11}\] \[t=\sqrt{\frac{5.4}{11}}s\]
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