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Kinematics ball problem

  1. Jan 21, 2007 #1
    I know I should be able to do this problem but for some reason it's not clicking with me...please help!

    1. The problem statement, all variables and given/known data
    a ball is thrown vertically upward with an initial speed of 19 m/s. Then, .97 s later, a stone is thrown straight up (from the same inital height as the ball) with an inital speed of 31 m/s. How far above the release point will the ball and stone pass each other. Answer is units of m. Gravity is -9.8 m/s^2.

    2. Relevant equations
    I tried using v^2=v0^2*2ax and x=x0+v0t+0.5at^2

    3. The attempt at a solution
    I solved 19(.97)+0.5(-9.8)(.97)^2 and got 13.82 m for the ball after .97 s.

    After this I got lost and didn't know what to do. I know this means that when the ball is at 13.82 m/s that the stone is just being thrown. Please help!!
  2. jcsd
  3. Jan 22, 2007 #2
    Do you have a graphing calculator? If so that might be a handy visualization tool.

    You are going to have two kinematic equations with distance as a function of time, and you want where they intersect. You will will probably end up with two answers, which means that one won't make sense and should be thrown out.
  4. Jan 22, 2007 #3
    I used to have one but I have no clue where it is at anymore. We aren't aloud to have them in this class so I have kinda lost track of it while I was moving.
  5. Jan 22, 2007 #4
    Well, your first equation looks okay if you put the time parameter back in.

    What does the second one look like?
  6. Jan 22, 2007 #5
    I actually figured it out today right after class (guess it got me in the right type of mode to do it). Sometimes I just don't look at them the way I should be. But what I ended up doing is plugging in 19 m/s for v, -9.8 for g (a) and 0.97 in for t into the equation
    v = v0 + at
    This gave me the answer of the velocity being 9.494 m/s when the ball is at 0.97 s and 13.82 m.
    Next, I plugged these numbers into the eqation
    x = x0+ v0t + 0.5at^2
    and set this equal to the other side. this ended up looking like
    13.82 + 9.494t + 0 = 0+ 31t+ 0
    solving for t I got 0.643 s.

    finally, I plugged all these numbers I solved for into
    x = x0 + v0t +0.5at^2
    this looked like
    13.82 + 9.494 (0.643) + 0.5 (-9.8) (0.643)^2
    The final answer ended up being 17.9 m.

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