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Free falling ball

  1. Oct 7, 2006 #1
    A ball bearing is dropped from rest at point A. At the instant it passes mark 3.5m below A, another ball bearing is realeased from rest from a position of 4.5m below A. At what time after its release will the second ball bearing be overtaken by the first?

    How far does the second ball bearing fall in that time?

    I know i have 2 distances and it starts at rest so it doesn't have an intial velocity i also know all falling objects have an acceleration of -9.8 so i used the formula d=1/2a(t)^2 and i used 3.5 for d -9.8 for a and t is my unknown so what am i doing wrong?
  2. jcsd
  3. Oct 7, 2006 #2
    first thing you want to do is find the velocity of the first ball as it hits the 3.5m mark. To do that...

    Vf^2 = Vi^2 + 2a(Xf-Xi) [Vf = final velocity; Vi = Initial velocity; (xf-xi) = displacement]
    Vf^2 = 0 + 2(9.8 m/s^2)(3.5m - 0) will use the initial position as A
    Vf = 8.282 m/s

    So we know that the first ball is traveling at 8.282 m/s as soon as the 2nd ball is being dropped. Now consider your distance formulas.

    Xf = Xi + ViT + 1/2aT^2

    Xf = 3.5m + (8.282 m/s)T + 1/2(9.8 m/s^2)(T^2) - equation for first ball
    Xf = 4.5m + (0) + 1/2(9.8 m/s^2)T^2 - equation for second ball

    Now we have two equations and two unknowns (time and position). The time and position in both equations are the same, so a simple subtraction will give us the answer.

    Xf = 3.5m + (8.282 m/s)T + 1/2(9.8 m/s^2)(T^2)
    Xf = 4.5m + (0) + 1/2(9.8 m/s^2)T^2
    0 = (-1m) + (8.282 m/s)T + 0
    T = 1/8.182 m/s
    T = .12 seconds

    Now that you have the time you can figure out how far the second ball falls.

    D = 1/2at^2 (this is only true when the ball is dropped from rest)
    D = 1/2(9.8 m/s^2)(.12)^2
    D = .070m
  4. Oct 7, 2006 #3
    ok i get everything but what xf stands for and thanks for the help!
  5. Oct 7, 2006 #4
    Xf = final position. Both balls will have the same final position so you can set their equations equal to each other (or subtract them)

    The equation you use D = 1/2at^2 is similar

    Xf = Xi + ViT + 1/2at^2 in my equation when you set the initial position to 0 and the initial velocity to zero you get...

    Xf = 1/2at^2 which is the same thing as your equation.

    Next time try to include all equations you believe have relevance to the problem so that I don't have to use my own notation which can be confusing.
  6. Oct 7, 2006 #5
    Thank you for the help i got it !!!!!
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