Question

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 covered by the bolt during the free fall time w.r.t. ground frame.

Answer 1

Explanation:

Let H be the height of the building and t be the time to fall through this height H.

Then H = ½ g t^2 = 5 t^2. ----------------- (1)

In (t-1) second the distance traveled is 1 - 0.36 H = 0.64 H.

0.64 H = 5 (t-1)^2.---------------------------(2)

(2) / (1) gives 0.64 = (t-1)^2/ t^2

Or 0.8 = (t-1) / t

Solving we get t= 5 second.

Using (1) we get H = 125 m.

Answer 2

Answer:

we get height =5m.........is ur answer....dude...