An object dropped in a stationary lift takes time t_1 to reach the floor. It takes time t_2 when lift is moving up with constant acceleration. Then

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Accepted Answer

Correct answer: B. (b) s=0.t_1+1/2gt_1 ^2 ⇒s= 12gt_1^2 … (A) [ ∵a=g ] If a_ net is net acceleration of ball, then when lift is going up with acceleration a a_ net=(g+a) ⇒s= 12(g+a)t_2^2 … … … (B ) From eqs. (A) and (B), we get t_1>t_2

$An object dropped in a stationary lift takes time t_{1} to reach the floor. It takes time t_{2} when lift is moving up with constant acceleration. Then$

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