Question

ig pelod. : An elevator accelerates from the ground with a uniform ac- celeration a. After 3 s, an object is dropped out of an opening in the floor of the elevator and that object hits the ground 3.5 s later. How large is the acceleration? How high was the elevator when the object was dropped? 6 .

Answer 1

For the falling object only force acting is the gravity.

So it took 3.5 s to hit the ground.

Let the elevator is at a height H from the ground,

$H=ut+\frac{1}{2}gt^{2}$

Here u is the initial velocty of the falling object, which is zero here.

$H=0+\frac{1}{2}\times9.8\times3.5^{2}=60.02\,m$

The elevator reached at a height H in 3s of time

$H=\frac{1}{2}\times a\times3^{2}=60.02$

$a=13.33\,m/s^{2}$