A skier leaves the ramp of a ski jump with a velocity of y = 12.0 m/s at theta =

Question

A skier leaves the ramp of a ski jump with a velocity of y = 12.0 m/s at theta = 19.0 degree above the horizontal, as shown in the figure. The slope where she will land is inclined downward at phi = 50.0 degree, and air resistance is negligible. (a) Find the distance from the end of the ramp to where the jumper lands. (b) Find the velocity components just before the landing. (Let the positive x direction be to the right and the positive y direction be up.) (c) Explain how you think the results might be affected ?f air resistance were included? This answer has not been graded yet.

Explanation / Answer

R = 2 u2 sin(alpha - beta) cos alpha / g cos2 beta

= 2 * 122 sin(19 - (-50)) cos 19 / 9.8 cos2 50

= 62 m

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t =  2 u sin(alpha - beta) / g cos beta

=  2 * 12 sin(69) / 9.8 cos 50 = 3.56 s

vx = ux = 12 * cos19 = 11.34 m/s

vy = uy - g t = 12 * sin19 - 9.8 * 3.56 = -30.8 m/s

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c) The presence of interaction with the air makes the competition more exciting. The skiers can influence the length of the jump. It would be surprising but theoretically jumps can be longer in the presence of interaction with the air! The idea is to reduce the y-component of the acceleration by using air as a source of an additional force (lift), while maintaining almost constant horizontal component of the velocity. The principle is applicable in hand gliding (when there is no air convection).

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