A skier leaves the ramp of a ski jump with a velocity of v = 26.0 m/s at theta =

Question

A skier leaves the ramp of a ski jump with a velocity of v = 26.0 m/s at theta = 18.0 degree above the horizontal, as shown in the figure. The slope where she will and is inclined downward at phi = 50.0 degree, and air resistance is negligible. Find the distance from the end of the ramp to where the jumper lands. Find the velocity components just before the landing. (Let the positive x direction be to the right and the positive y direction be up.) Explain how you think the results might be affected if air resistance were included?

Explanation / Answer

From the given data:
velocity = 26m/s
angle = 18 degrees
inclination = 50 degrees
Use the trajectory equation: y = h + x·tan - g·x² / (2v²·cos²)
where y = height at given value of x; here y = -x*tan50
and h = initial height = 0 m
and x = range = ???
and = launch angle = 18 degrees
and v = 26 m/s

-xtan50 = 0 + xtan18 - 9.8x^2 / (2(26)^2*(cos(17))^2)
0 = -x*2.719 - 1.1373*x -x^2*0.9574
0=3.8563+0.9574 x
x=-4.0278 or
so either x = 0 (trivial)

y = 4.0278*tan50 = 1.095 m

(a) d = (x² + y²) = 4.174 m

(b) Vx = Vcos18 = 17.16 m/s
Vy = ((Vsin)² - 2as) = ((26sin18)^2 + 2*9.8*1.095) m/s = 20.067 m/s

(c) Both velocities would have smaller magnitude, since drag would retard motion in both directions.

(d) Because it increases their drag coefficient.

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