# Homework Help: Ball rolling in elevator

1. Jul 25, 2015

### Fuzzykatecake

An open elevator is moving with an upward velocity of
0.52ms-1 with an upward acceleration of 2.4ms-2. A ball bearing then rolls off the floor with zero horizontal speed. Determine the speeds of ball bearing and elevator 2s after the ball rolls off the floor.

Using v=u+at, speed of ball bearing
v=(-0.52)+(9.81)(2)

I don't understand why the acceleration of ball is 9.81 and not acceleration of lift. Shouldn't the acceleration of ball be the same as lift as they are in contact?

Last edited: Jul 25, 2015
2. Jul 25, 2015

### haruspex

What do you think 'flier' refers to? (This is a genuine question, I don't know what it means here.) My guess is it means it rolls out of the elevator.

3. Jul 25, 2015

### Fuzzykatecake

I am so sorry! It's floor!

4. Jul 25, 2015

### haruspex

So does that resolve your question?

5. Jul 25, 2015

### Fuzzykatecake

Wait.. Are u saying that the ball rolls OUT of the elevator and is free falling and that's why the a is 9.81. But it still has the upward velocity of lift.. But how is it possible?

6. Jul 25, 2015

### Staff: Mentor

When it rolls out of the lift, it is initially still traveling with an upward velocity of 0.52 m/s. It is then subjected to a downward acceleration of 9.8 m/s2.

Chet

7. Jul 25, 2015

### Fuzzykatecake

Now I get it thanks:)