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(C) The elevator is moving upwards with an acceleration $a = 1.2 \, m/s^2$. When the bolt detaches from the ceiling,it is in a frame of reference acce

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$0.7$

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(C) The elevator is moving upwards with an acceleration $a = 1.2 \, m/s^2$. When the bolt detaches from the ceiling,it is in a frame of reference accelerating upwards.The effective acceleration of the bolt relative to the elevator is $g_{eff} = g + a$,where $g = 9.8 \, m/s^2$ is the acceleration due to gravity.Given the height $h = 2.7 \, m$,the time $t$ taken for the bolt to reach the floor is given by the kinematic equation $h = \frac{1}{2} g_{eff} t^2$.Substituting the values:$2.7 = \frac{1}{2} (9.8 + 1.2) t^2$$2.7 = \frac{1}{2} (11) t^2$$2.7 = 5.5 t^2$$t^2 = \frac{2.7}{5.5} \approx 0.49$$t = \sqrt{0.49} = 0.7 \, s$.

An elevator car,whose floor-to-ceiling distance is $2.7 \, m$,starts ascending with a constant acceleration of $1.2 \, m/s^2$. $2 \, s$ after the start,a bolt begins falling from the ceiling of the car. The free-fall time of the bolt is ......... $s$.

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