An elevator, whose floor to ceiling distance is equal to 2.7m, starts

1.	An elevator, whose floor to ceiling distance is equal to 2.7m, starts ascending with constant acceleration of 1.2m/s^2. 2 sec after the start, a bolt begins falling from the ceiling of the car. The free fall time of the bolt is
Answer» Here, h = 2.7 m a1 = 1.2 ms-2 g = 9.8 ms-2 u = 0 (w.r.t. elevator) Net acceleration of the bolt, as the elevator is moving up with acceleration a, Will be a = a1 +g = 11 ms-2 Let time at which the ball starts dropping be t = 0 h = ½ at2 => t = √(2h/a) => t = √(2×2.7/11) =>t = 0.7 s Now time when it reaches the floor from starting of the elevator will be = 2+ 0.7 = 2.7 s

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

Answer»

Here,

h = 2.7 m

a1 = 1.2 ms-2

g = 9.8 ms-2

u = 0 (w.r.t. elevator)

Net acceleration of the bolt, as the elevator is moving up with acceleration a,

Will be

a = a1 +g = 11 ms-2

Let time at which the ball starts dropping be t = 0

h = ½ at2

=> t = √(2h/a)

=> t = √(2×2.7/11)

=>t = 0.7 s

Now time when it reaches the floor from starting of the elevator will be = 2+ 0.7 = 2.7 s

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

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