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Constant Acceleration of a skier Problem

  1. Jan 22, 2013 #1
    1. The problem statement, all variables and given/known data

    A skier travels down a slope and leaves the ski track moving in the horizontal direction with a speed of 28 m/s. The landing falls off with a slope of 32 degrees.
    a.) How long is the skier airborne? ignoring air resistance
    b.) How far down the incline does the jumper land along the incline?
    2. Relevant equations



    3. The attempt at a solution
    a.) Since V_x*t=d_x I was going to find how far down the incline the skier lands then solve for two but the order of the questions makes it seem that that is how you solve part. Finding the t value first.

    I wasn't really sure where to go with this but maybe finding the interesting point of the slope and the parabolic trajectory of the skier but the only information I have is the slope of the incline so I don't know how to set that up..any help?
     
  2. jcsd
  3. Jan 22, 2013 #2

    Simon Bridge

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    This is a ballistics problem - I always tell people to draw the v-t diagrams.
    In this case you have a bit of an issue in that the distance fallen also depends on the horizontal distance covered. You want to find where the parabola of the trajectory intersects the line of the landing.

    You will end up with two equations and two unknowns - you can solve for them in either order.
     
  4. Jan 22, 2013 #3
    I understand that I need to find where they intersect but I don't know how get the equations of the parabola and the line. I have the slope of the incline but no point. And I don't know the parabola either...how do I come up with the equations?
     
  5. Jan 22, 2013 #4

    rude man

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    Set up a coordinate system with (x=0,y=0) the point, and t=0 the time, he leaves the track.

    Questions for you to answer:
    1. What is his x(t)?
    2. What is his y(t)?
    3.What is the relationship between x and y for the slope?

    I would then solve for the x value of the impact point, from which the answers follow immediately from the above equations.
     
  6. Jan 22, 2013 #5
    ok thank you!
     
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