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Position, velocity

  1. Dec 20, 2005 #1
    if a person throws a rock strait up with an initial velocity of 25 m/s and a rock is dropped from a height of 50 ft. how do you determine the time at which they will meet in the air?:grumpy:
  2. jcsd
  3. Dec 20, 2005 #2
    You find two equations which describe the position of the two rocks as a function of time and then set them equal to one another.

    If you have any more questions please post back with whatever work you have attempted so that we have a better idea as to how we can help you.
  4. Dec 20, 2005 #3
    i got x = vi + (at^2)/2 and i plugged in my information and set them equal to eachother but time comes out negative :bugeye:
  5. Dec 20, 2005 #4
    Up: [tex]y_{2} = y_{1} + v_{1y}t + \frac{1}{2}at^2[/tex] v1 = 25, a = -9.8, what's your initial height when you throw it straight up? from the ground?

    Drop: [tex]y_{2} = y_{1} + v_{1y}t + \frac{1}{2}at^2[/tex] v1=0, a=-9.8, y1=50.

    You could set the y2's equal to each other.

    [tex]50 - \frac{1}{2}gt^2 = 25t - \frac{1}{2}gt^2[/tex]
    [tex]50 = 25t[/tex]

    So, t=2 if the rock is thrown up from the ground, and if you take t=2 and put it into each, you end up with the same height, 30.4m.
  6. Dec 20, 2005 #5
    u have 4 motion equations to work from.
  7. Dec 22, 2005 #6


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    Maybe it came out negative cause you forgot to put a minus sign on the acceleration of the rock going up?
  8. Dec 24, 2005 #7
    Maybe you should show some work instead of being a lazy person and having others do it for you?
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