Ball Bearing Speed Calculation in Elevator

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Fuzzykatecake
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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
Answer given was
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?
 
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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.
 
haruspex said:
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.
I am so sorry! It's floor!
 
Fuzzykatecake said:
I am so sorry! It's floor!
So does that resolve your question?
 
haruspex said:
So does that resolve your question?
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?
 
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
 
Now I get it thanks:)