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Projectile motion of a skier

  1. Jan 31, 2012 #1
    1. The problem statement, all variables and given/known data
    A skier leaves the ramp of a ski jump with a velocity of v = 13.0 m/s, θ = 15.0° above the horizontal, as shown in the figure. The slope where she will land is inclined downward at = 50.0°, and air resistance is negligible.


    2. Relevant equations



    3. The attempt at a solution
    I thought the answer was 70.1, but I dont think that im doing the correct process
     
  2. jcsd
  3. Jan 31, 2012 #2
    Where's the figure?
     
  4. Jan 31, 2012 #3
    If you are finding the final horizontal velocity:
    13cos15 = vcos50

    If you are finding the final vertical velocity:
    13sin15 - 9.8t = vsin 50

    If you are finding the velocity, it is equal to the square root of the sum of the squares of the above two results. The direction is given by tan^-1(vx/vy).
     
  5. Jan 31, 2012 #4
    Are you asking for the distance from the jump to the slope? Or the time from the jump to the slope?

    Either way, you need to set up the equation [13.0 m/s*t*cos(15.0)] (i component) + [13.0 m/s*t*sin(15.0) - 1/2*g*t^2] (j component)

    You can then solve for time when you divide the scalar portion of the components and set them equal to tan(-50.0) = (j component)/(i component)

    Assuming g = 9.8m/s^2
    t = 2.36s
     
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