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Ball in Elevator

  1. Apr 3, 2005 #1
    Ball in Elevator......

    Q.
    Ball in Elevator

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    A physicist is inside an elevator rising in a skyscraper. She is holding a ball at a height h = 2.1 m above the floor of the elevator.

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    a) She lets go of the ball while the elevator is rising at a constant speed of 4.6 m/s. How long does it take the ball to hit the elevator floor?
    tto floor = s
    2.1/4.6 NO

    HELP: The ball and the elevator floor will be at the same height above the ground when the ball hits the floor

    I used

    x - x0 =v0t + 1/2at^2

    Since constant speed acc. is zero.

    so,

    t = x-xo/vo
    2.1 /4.6 which is wrong???

    Any one help
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    b) Now let's try a somewhat harder variation on this problem. Suppose instead that the elevator is accelerating upward with constant acceleration a = 1.2 m/s2 and that the upward speed of the elevator at the instant the ball is dropped, is 4.6 m/s. Now how long does it take the ball to hit the elevator floor, assuming once again that it is dropped from a height of 2.1 m? Q


    No idea here help me, what equations, may be the same one as above, what to plugin, etc.etc.
     
  2. jcsd
  3. Apr 4, 2005 #2
    The acceleration of the elevator is zero, the acceleration due to gravity is still 9.8 m/s. For both parts you need to write separate position equations for both the elevator and ball, and solve for when they are equal.
     
  4. Apr 4, 2005 #3
    The ball is falling to the floor of the elevator and the elevator is rising to meet the ball. Think of two particles moving towards each other. Write their positions over time and solve them simultaneously to find when they intersect.
    For the ball

    [itex]y = y_0 -\frac{1}{2}gt^2[/itex]

    like you said. For the floor of the elevator

    [itex]y_{\mathrm{fl}} = y_{\mathrm{fl},0}+ v_{\mathrm{fl}} t.[/itex]

    Same thing as above, except you need to include the [itex]1/2 at^2[/itex] term in the second equation to take into account the elevator's upward acceleration.
     
  5. Apr 5, 2005 #4
    Yep! Got em all! Thanks,
     
  6. Apr 6, 2005 #5
    hello! I have the same problem only with different values:

    the height of the ball is h = 1.2 m instead.

    a) She lets go of the ball while the elevator is rising at a constant speed of 4.7 m/s. How long does it take the ball to hit the elevator floor?

    what i did was as listed above i did:

    [itex] y = 1.2 - \frac{1}{2}(9.8)t^2[/itex] for the ball and

    [itex]y_{\mathrm{fl}} = 0 + 4.7t[/itex] for the elevator

    i graphed those two and the intersection is at t = .2095

    why is this not the right answer? thanks!
     
    Last edited: Apr 6, 2005
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