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Kinematics-acceleration in two dimensions

  1. Sep 10, 2012 #1
    1. The problem statement, all variables and given/known data

    A tennis player standing 8.0 meters from the net hits a ball 1.5 meters above the ground toward her opponent. The ball leaves her racquet with a speed of 25.0m/s at an angle of 14.0 degrees above the horizontal. The net is 1.0 meters high. The baseline is 12 meters back from the net; a ball striking the ground beyond the baseline is out of play

    2. Relevant equations
    x=V.costheta(t)
    y=V.sintheta(t)-0.5g(t^2)


    3. The attempt at a solution

    I attempted to solve for t using the quadratic equation in conjunction with the above formula for y. I ended up with two solutions, neither was extraneous from what I could tell. I plugged both in for x and ended up with two answers. One was slightly more than 8 (meaning that the ball landed in), the other was a little over 21 (meaning that the ball landed out). I am lost, which is it and how do I know?
     
  2. jcsd
  3. Sep 10, 2012 #2

    gneill

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    Staff: Mentor

    You need to include the initial height of the ball (where its launched) when you work with the vertical component of the position.
     
  4. Sep 10, 2012 #3
    Right, and I did. I got the equation deltaY=V.sintheta(t)-0.5gt^2

    Rearranged and solve the quadratic equation. I ended up with two answers. One was 0.89 and the other was 0.3437. How am I supposed to determine which is correct and which is not?
     
  5. Sep 10, 2012 #4

    gneill

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    Staff: Mentor

    I'll bet you've set your Δy to a positive value, right :wink: But doesn't it FALL from its initial position to hit the ground?
     
  6. Sep 10, 2012 #5
    :facepalm: I appreciate the help. I remember running into this same problem in high school. Maybe after learning from the same mistake twice, I won't make it again. Fingers crossed.
     
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