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Highland Games hay bale toss!

  1. Sep 15, 2011 #1
    2D Kinematic Equation! Need Help! I have a test tomorrow!

    1. The problem statement, all variables and given/known data
    One of the most popular events at highland games is the hay toss, where competitors use a pitchfork to throw a bale of hay over a raised bar. Suppose the initial velocity of a bale of hay is varrowbold = (1.32 m/s)xhatbold + (8.85 m/s)yhatbold.
    (a) After what minimum time is its speed equal to 7 m/s?

    (b) How long after the hay is tossed is it moving in a direction that is 45.0° below the horizontal?





    2. Relevant equations
    I figured out the answer to part A using the quadriatic equation by saying that ...
    t=2Vnotyg+-sqrt of (-2Vnotyg)^2 - 4(g^2)(Vnotx^2+Vnoty^2-V^2)/2(g^2).
    I came up with the right answer of .201s.

    I am stuck with part B. I don't know if I am overanalizing it or what. Someone please help!
     
    Last edited: Sep 15, 2011
  2. jcsd
  3. Sep 15, 2011 #2

    PeterO

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    Re: 2D Kinematic Equation! Need Help! I have a test tomorrow!

    When it is moving at 45 degrees, the verticla (y) and horizontal (x) component are the same size.
    At 45 degrees down means the y-component is 1.32 down.
     
  4. Sep 15, 2011 #3
    So I would say 1.32sin45, which would equal what? I am confused ... My mind is scrambled right now.
     
  5. Sep 15, 2011 #4

    PeterO

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    The original velocities given didn't have a sin45 component, so why did you put one in.

    Hint: at what time will the hay bale have a vertical component of 5 m/s down? If you can work that out, you can solve part (b) in a similar way
     
  6. Sep 15, 2011 #5
    So I would use the Vy=Voy+Ayt right? And since Voy=0, the equation would be t=Vy/Ay?
     
  7. Sep 15, 2011 #6
    Nm, I got it! It was t=x+y/g! I am over analizing this stuff way too much!
     
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