Question

An elevator car whose floor to ceiling distance is equal to 2.7 m starts ascending 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) The velocity of bolt at instant it loses contact is

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:

$\bf\huge\underline\red{Akash}$

Explanation:

is correct one........