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Can someone help me with this problem. Two stones being thrown straight up

  1. Sep 14, 2006 #1
    A stone is thrown vertically upward at a speed of 18.00 m/s at time t=0. A second stone is thrown upward with the same speed 0.880 s later.
    (a) At what time are the two stones at the same height?
    (b) At what height do the two stones pass each other?
    (c) What is the upward speed of the second stone as they pass each other?
    (d) What is the downward speed of the first stone as they pass each other?

    I'm at a complete loss for this. I know stone one will be traveling down and stone two will be traveling up when they are at the same height, but how do I find that exact height.

    Here are the formulas I'm allowed to work with.
    Xf = Xi + VT
    Vf = Vi + AT
    Xf = Xi + ViT + 1/2AT^2
    Vf^2 = Vi^2 + 2AXf

    BTW I don't need the answer to this exact problem. I just need to know how to do it on my test. Any help will be greatly appreciated.
  2. jcsd
  3. Sep 14, 2006 #2


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    Welcome to the Forums,

    Start by listing the things you know;

    (a) You know the initial velocity of both stones (18 m.s-1)
    (b) You know the acceleration of both stones (-9.81 m.s-2)
    (c) You know that when both stones are at the same height, they will have equal displacements.

    You want to find out at what time t both stones are at the same height, so you need an equation with the initial velocity, the acceleration, the displacement and time only. Which equation from your list would that be?
  4. Sep 14, 2006 #3
    Would it be the 3rd equation? Even after I know the equation I'm still confused. This is my first attempt at any sort of physics and its really making me think.
  5. Sep 14, 2006 #4
    How do I know they will have equal displacements at the same height?
  6. Sep 14, 2006 #5


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    That's what it means to be at the same height. If one stone is 5 meters above the ground and the other stone is also 5 meters above the ground, they are both displaced from their point of origin (the ground) by 5 meters.

    Do you know the formula for displacement?
  7. Sep 14, 2006 #6
    No, I don't. Our class doesn't have a book so I only have what my prof gave us. I get displacement now, I thought it was the total distance traveled before you just explained it.
  8. Sep 14, 2006 #7


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    Ask yourself, what is height? What is displacement? How could the stones be at the same height and not have the same displacement?
    Yes, it would be the third equation. Now, can you show two equations (one for each stone) and substitute in the values that you know?
  9. Sep 14, 2006 #8
    Do I make the two equations equal to each other?

    Xi + ViT + 1/2AT^2 = Xi + ViT + 1/2AT^2 ?

    16.53m + 0m/s(T) + .5(-9.8m/s^2)(T^2) = 0m + 18m(T) + .5(-9.8m/s^2)(T^2)
  10. Sep 14, 2006 #9


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    Yes, you do make both equations 'equal', however I am not sure what you are doing here. Surely both stones have the same initial displacement (x1 = 0) and the same initial velocities (vi = 0). The only differences are that in the first case T = t and in the second case T = t + 0.88. Does that make sense?
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