0 0 A skier leaves the ramp of a ski jump with a velocity of v-11.0 m/s at 11.0°

Question

0 0 A skier leaves the ramp of a ski jump with a velocity of v-11.0 m/s at 11.0° above the horizontal, as shown in the figure. The slope where she will land is indíned downward at -50.0°, and air resistance is negligible. (a) Find the distance from the end of the ramp to where the jumper lands st before the landing. Let the positive x direction be to the right and the positive y direction be up.) m/s m/s (c) Explain how you think the results might be affected if air resistance were included? s answet has not been araded vet.

Explanation / Answer

Initial velocity, u = 11 m/s

X component of velocity, ux = 11 x cos11 = 10.80 m/s

Y-component of velocity, uy = 11 x sin11 = 2.10 m/s

Tan(-50) = - 1.19 = (2.10t - 4.9t^2)/(10.80t)

On solving, we get t = 3.05 sec

X2 - x0 = 10.80 x 3.05 = 32.96 m

So, down slope distance, d = 32.96/cos50 = 51.27 m

B)

Vxf = 10.80 m/s

Vyf = 2.10 - (9.8 x 3.05) = - 27.79 m/s

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