What Is the Distance a Skier Lands From a Ramp?

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A skier leaves a ramp at 10.0 m/s and 15.0° above the horizontal, with the ramp inclined at 50.0°. The equations used to calculate the landing distance include the vertical and horizontal motion equations. The user initially miscalculated the vertical final position (Yf) as positive instead of negative, leading to incorrect time and distance results. After recognizing this error, the user is seeking clarification on the correct approach to solve for the landing distance. Accurate calculations are crucial for determining the skier's landing point.
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Homework Statement


A skier leaves the ramp of a ski jump with a velocity of 10.0 m/s, 15.0° above the horizontal, as shown in Figure P3.57. The slope is inclined at 50.0°, and air resistance is negligible. Find the distance from the ramp to where the jumper lands .
image092020141179.png

Homework Equations


tan50degrees=Yf/Xf
Yf = Yi + Vyi(t) + .5(ay)t2
Xf=Xi+Vx(t)

The Attempt at a Solution


Plugging in numbers into the first second equation using the first equation and solving for distances gave me the following: Xf(tan50) = 2.59t + -4.9t2 . Then I solved for the third equation and resulted in Xf=9.66t. Plugging this into the partially solved second equation yielded: 9.66tan(50)t = 2.59t - 4.9t2 which simplifies to -4.9t2 - 8.91t. t is supposed to equal 2.88 seconds but my solution doesn't yield that at all. What am I doing wrong? Thanks.
 
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got it... the Yf i used was not negative.
 

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