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2-dimensional kinematics

  1. Sep 23, 2011 #1
    This is an even problem with no solution provided. My quiz is Monday. Could somebody check this practice problem for me?

    A skateboarder shoots off a ramp with a velocity of 6.6 m/s, directed at an angle of 58° above the horizontal. The end of the ramp is 1.2 m above the ground. Let the x axis be parallel to the ground, the +y direction be vertically upward, and take as the origin the point on the ground directly below the top of the ramp.


    (a) How high above the ground is the highest point that the skateboarder reaches?
    (b) When the skateboarder reaches the highest point, how far is this point horizontally from the end of the ramp?


    Vo = 6.6 m/s
    Voy = Vosin58 = 6.6sin58
    Vox = Vocos58 = 6.6cos58

    Vy = 0 m/s at the highest point

    Vy2 = Voy2 + 2ay
    y = ( Vy2 - Voy2 ) / 2a
    y = [0 - (6.6sin58)2] / (2*-9.8)
    y = 1.598353244 m

    y + 1.2 = 2.798 m

    VFy = 0 m/s
    Voy = 6.6sin58

    VFy = Voy + at
    t = ( VFy - Voy ) / a
    t = ( 0 - 6.6sin58 ) / -9.8
    t = 0.5711344321 s

    Vx = x/t
    x = Vxxt = (6.6cos58)(0.5711344321)
    x = 1.997 m
     
  2. jcsd
  3. Sep 23, 2011 #2

    Delphi51

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    Homework Helper

    They all agree with my answers.
     
  4. Sep 23, 2011 #3
    Thank you!!!!
     
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