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(B) When the object is dropped in a stationary lift from height $h$,the equation of motion is $s = ut + \frac{1}{2}at^2$.Since $u = 0$ and $a = -g$,we

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$t_1 > t_2$

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(B) When the object is dropped in a stationary lift from height $h$,the equation of motion is $s = ut + \frac{1}{2}at^2$.Since $u = 0$ and $a = -g$,we have $-h = -\frac{1}{2}gt_1^2$,which gives $t_1 = \sqrt{\frac{2h}{g}}$.When the object is dropped from a lift moving upward with constant acceleration $a$,the object experiences a pseudo force in the downward direction relative to the lift frame.The effective acceleration becomes $g_{eff} = g + a$.Using the same kinematic equation,$-h = -\frac{1}{2}(g + a)t_2^2$,which gives $t_2 = \sqrt{\frac{2h}{g + a}}$.Since $(g + a) > g$,it follows that $\frac{2h}{g + a} Therefore,$t_2  t_2$.

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

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