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Ball Dropping vs Throwing (1d Kinematics)

  1. Oct 10, 2006 #1
    A ball is thrown upward from the ground with an initial speed of 30 m/s; at the same instant, a ball is dropped from a building 20 m high. After how long will the balls be at the same height?

    Again, some guidance on where to start would help me crack this one better.
     
  2. jcsd
  3. Oct 10, 2006 #2
    [tex]y=y_0+v_0t+\frac {1} {2}at^2[/tex]
    Use this equation for both balls using the same zero level. Their Y levels will be the same so you can put both equations together and solve for t.
     
  4. Oct 10, 2006 #3
    Is y 20? and y0 = 0? I know that Vo is 0.
     
  5. Oct 10, 2006 #4
    y = the position at which the 2 balls meet
    y0 is the initial position. If you set the 0 level at the ground then the ground ball y0 = 0 and the building ball y0 = 20. The building ball will have v0 = 0, the ground one will have v0=30m/s. Make 2 different equations with that information, combine then, solve for T
     
  6. Oct 10, 2006 #5
    Building Ball y = 20 + 0t + 1/2(-9.81)t^2
    Ground Ball y = 0 + 30t + 1/2(-9.81)t^2

    Do I combine them by doing y = 20 + 30t + 1/2(-9.8)t^2
     
  7. Oct 10, 2006 #6
    both equations are equal to the same y, so you combine them by setting them equal to one another

    20 + 0t + 1/2(-9.81)t^2 = 0 + 30t + 1/2(-9.81)t^2
    20 = 30t
    t = 2/3 seconds
     
  8. Oct 10, 2006 #7
    Thank you, you learn something everyday :)
     
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