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An elevator car whose floor to ceiling distance is equal to `2.7 m` starts ascendiung with constant acceleration `1.2(m)/(s^(2))`, 2 sec after the start a bolt begins falling from the ceiling of the car. Answer the following question `g = 9.8 m//s^(2)` Distance moved by elevator car w.r.t. ground frame during the free fall time of the bolt.A. 1.44 mB. 1.63mC. 1.68mD. 1.97m |
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Answer» Correct Answer - D `y=ut+(1)/(2)at^(2)=(2.4)(0.7)+(1)/(2)(1.2)(0.7)^(2)` |
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