Question

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

Answer 1

Answer: Approximately 0.952 seconds



The bolt is initially at the ceiling of the elevator, which is 2.7 metres above the floor of the elevator.

The elevator starts ascending from rest. Before 2 seconds, the bolt is also attached to the ceiling. So, as lift accelerates upwards, the bolt also acquires an upward velocity, say $u_{\circ}$


Let us calculate this upward velocity just before the bolt is released.


$v=u+at \\ \\ \implies u_{\circ} = 0 + (1.2)(2) \\ \\ \implies u_{\circ} = 2.4 \, \, m/s$

So, just before the bolt is released, it has an upward velocity of 2.4 m/s.

Now, the bolt is released. Let us observe the situation from the ground frame. The bolt is falling towards the floor with an acceleration of 9.8 $m/s^2$

However, the lift itself is moving upwards. The floor is also moving towards the bolt with an acceleration of 1.2 $m/s^2$

Now, consider the Frame of the Bolt. The bolt is moving towards the floor with a relative acceleration of:

$a_{rel} = 9.8+1.2 = 11 \, \, m/s^2$

So, the relative acceleration of the bolt with respect to the floor is 11 $m/s^2$ downwards. Not to forget that the bolt had an upward velocity of 2.4 m/s when it started free-fall. 

Let the time of free fall be $t$. By Equations of Motion, we can find the time.

Initial Velocity is an upward 2.4 m/s
Acceleration is a downward 11 $m/s^2$
Displacement is 2.7 metres downward.

So, by considering upwards as positive, and downwards as negative, we have:

$s=ut+\frac{1}{2}at^2 \\ \\ \implies -2.7 = (2.4)(t) + \frac{1}{2} (-11)(t^2) \\ \\ \implies -2.7 = 2.4t - 5.5t^2 \\ \\ \implies 5.5t^2-2.4t-2.7=0 \\ \\ \text{Multiplying both sides by 10} \\ \\ \implies 55t^2-24t-27=0\times 100 \\ \\ \implies 55t^2-24t-27=0$

So, the solution is:



$t = \frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ \\ \\ \implies t = \frac{24\pm \sqrt{(-24)^2-4(55)(-27)}}{2(55)} \\ \\ \\ \text{Rejecting negative solution for time} \\ \\ \\ \implies t = \frac{24+\sqrt{576+5940}}{110} \\ \\ \\ \implies t = \frac{24+\sqrt{6516}}{110} \\ \\ \\ \implies \boxed{t \approx 0.952 \, \, s}$



Thus, The Free-Fall Time of the Bolt is approximately 0.952 seconds.

Answer 2

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.