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Physics Kinematics Assistance

  1. Oct 10, 2009 #1
    1. A bike rider leaves a ramp at an angle of 32 degrees and lands at the same height after traveling a horizontal distance of 36m. What was his take off speed?

    2. S=ut+1/2at^2

    3. My attempt:
    Vertical : s u v a=9.81 t=?
    Horizontal: s=36m u v a=0 t=?

    36/(xcos32) = t

    xsin32/4.905 = t

    36/xcos32 = xsin32/4.905

    36*4.905 (xsin32)(xcos32)

    Thats my best attempt, hopefully someone can tell me if I am going wrong and what is the next step in order to complete the question. Thanks
  2. jcsd
  3. Oct 10, 2009 #2


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    Homework Helper

    36*4.905 = (xsin32)(xcos32) looks okay to me.
    To solve for x, divide both sides by sin(32)*cos(32)
    and take the square root of both sides.
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