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Problem about Free Falling Objects

  1. Oct 19, 2012 #1
    1. The problem statement, all variables and given/known data

    A ball is thrown vertically upward with an initial speed of 20 m/s. Two seconds later, a stone is thrown vertically (from the same initial height as the ball) with an initial speed of 24 m/s. At what height above the release point will the ball and stone pass each other?


    2. Relevant equations

    yf = yi + vit + ½ at2


    3. The attempt at a solution

    The right solution :

    The ball is thrown 2 sec before the rock
    yball = ½ at2 + vit + yi
    yball = ½(-10)(t+2)2 + 20(t+2)
    yball = -5t2 + 20

    yrock = ½ at2+ vit + yi
    yrock = ½(-10)t2 + 24t
    yrock = -5t2 + 24t

    yball = yrock (they pass each other)
    -5t2 + 20 = -5t2 + 24t
    t = 5/6 seconds

    yrock = -5(5/6)2 + 24(5/6)
    yrock = 17 meters

    but I don't know why he considered (t) of the ball (t+2) althought the ball is thrown 2 seconds before the stone???
    isn't it true if :
    t of the ball = t-2
    t of the stone = t
    ?
     
    Last edited: Oct 19, 2012
  2. jcsd
  3. Oct 19, 2012 #2
    Where is the ball at t=0 and where is the rock at t=0? The reason you use t+2 instead of t-2 is because the equation you used for the rock assumed it was y=0 at t=0, but it was at y=0 at t=-2. t-(-2)=t+2
     
  4. Oct 19, 2012 #3
    Thank you. that helps.
    but I will be pleased with more clarification.
     
  5. Oct 19, 2012 #4
    The equation you used both times, [itex]y=v_0t-\frac{1}{2}gt^2[/itex] is for a projectile starting at position 0m. You have to have t the same for both objects, and the ball is not at its starting position of 0m at t=0s, but it was at t=-2s.
     
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