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Kinematics / forces - Young girl swinging on a rope

  1. Feb 2, 2006 #1
    On a hot summer day a young girl swings on a rope above the local swimming hole. When she lets go of the rope her initial velocity is 2.65 m/s at an angle of 35.0° above the horizontal. If she is in flight for 1.30 s, how high above the water was she when she let go of the rope?

    can anyone help?? thanks.
     
  2. jcsd
  3. Feb 2, 2006 #2

    KD

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    To solve this problem, it is best to set up a chart and draw a picture.
    We already know the time is 1.30s and the accleration is 9.81 m/s2 - that's always a given. To get the vertical height, you need a vertical initial velocity. You need to break the velocity they give you into components. Make a right triangle and put that 2.65 at the hypotenuse. Solve for the vertical inital velocity by 2.65sin35.
    After that you can plug it into the equation d=vi t +.5 a t^2
    This will give you the height.
     
  4. Feb 2, 2006 #3

    lightgrav

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    Homework Helper

    First separate the velocity vector into vertical and horizontal components.
    Then use the (known) vertical acceleration and the (known) time
    to compute the change in vertical location.
     
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