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Ball rolling in elevator

  1. Jul 25, 2015 #1
    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?
     
    Last edited: Jul 25, 2015
  2. jcsd
  3. Jul 25, 2015 #2

    haruspex

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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.
     
  4. Jul 25, 2015 #3
    I am so sorry! It's floor!
     
  5. Jul 25, 2015 #4

    haruspex

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    So does that resolve your question?
     
  6. Jul 25, 2015 #5
    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?
     
  7. Jul 25, 2015 #6
    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
     
  8. Jul 25, 2015 #7
    Now I get it thanks:)
     
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