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Projectile Motion - Skier & Ramp

  1. Jun 16, 2015 #1
    1. The problem statement, all variables and given/known data
    A ski jumper acquires a speed of 119.0 km/hr by racing down a steep hill. He then lifts off into the air from a horizontal ramp. Beyond this ramp, the ground slopes downward at an angle of θ = 45 degrees.

    2. Relevant equations
    Assuming the skier is in free-fall motion after he leaves the ramp, at what distance d down the slope does the skier land?

    3. The attempt at a solution
    I converted 119.0 km/hr to m/s, so 33.055 m/s

    I then try to find the horizontal range: 0=33.055^2 + 2*9.8*X and found X=55.74m
    55.74cos45=78.83m down the ramp.

    What have I done wrong?
  2. jcsd
  3. Jun 16, 2015 #2


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    That formula is for taking off and landing at the same height.
    Go back to the SUVAT equations and solve from first principles.
  4. Jun 17, 2015 #3
    With respect to the slope which is angled at 45 dgree, the ski jumper is jumping off the ramp with a velocity u=(119x5/18)m/s at angle of 45 degree!
    Thus, the horizontal range he covers on the slope, d=(u^2xsin(2θ)/g).
    Correct me if im wrong,or you can ask for further doubts or queries.
  5. Jun 17, 2015 #4


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    You are wrong. Horizontal means horizontal - it doesn't mean parallel to the slope.
    Gravity continues to act vertically, not orthogonally to the slope.
  6. Jun 17, 2015 #5
    Thanks for the glitch. :oldsmile:
    Ok, suppose 'd' is the hypotenuse of an isosceles triangle of angle 45 degree and its equal sides=dcos45 or (dsin45 whatever). Then time taken for the skii jumper to cover 'd' is t=(2xh/g)½ = (2xdcos45/g)½.
    Now, R=dsin45=u x t= (u^2 x 2 x dcos45/g)½
    solving out the above equation will give the value of 'd'.
  7. Jun 17, 2015 #6


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    Yes it's correct now but you should let the original poster figure it out for themselves so they can learn too :oldsmile:
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