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BASIC physics question: baseball going at Vo

  1. Nov 9, 2011 #1
    1. The problem statement, all variables and given/known data
    A baseball is hit at a height of 1m with an initial speed of vo module at an angle of projection of 35o from the horizontal. She spent just over an obstacle of 29m high at a horizontal distance of 64m.
    Find
    a) vo
    b) the time it takes to reach the OBSTACLES
    c) its speed at the obstacle.


    2. Relevant equations
    Xi= 0
    Xf= 64
    Yi= 1
    Yf= 29

    Equations: Yf= Vosin(35) -9,8t

    3. The attempt at a solution
    Not too sure how to go about with this one. I end up with two variables in the equations i use. help!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 9, 2011 #2

    Andrew Mason

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    Start with:

    [itex]v_x = v_0\cos(\theta)[/itex]

    [itex]v_y = v_0\sin(\theta) - gt[/itex]

    What are the equations for the height and range as a function of time?

    Set height at 29 m and at horizontal range 64 m. Plug the values for x, y into these equations. From that you should be able to determine v_0.

    AM
     
  4. Nov 14, 2011 #3
    Sorry it took a while, heres what I did... something's off...
    just part a)

    A7543C0D.jpg
     
  5. Nov 14, 2011 #4

    Andrew Mason

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    From the first equation:

    [itex]t = 64/v_0\cos{(35)}[/itex]

    Substitute that value for t into the second equation:

    [itex]y = y_0 + v_{0}\sin{(35)}t - \frac{1}{2}gt^2[/itex]

    It gets a little hairy with a quadratic equation but I think it is a little easier to solve for v0 than for t.

    AM
     
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