What Is the Distance a Skier Lands From a Ramp?

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SUMMARY

The discussion centers on calculating the distance a skier lands from a ramp, given an initial velocity of 10.0 m/s at an angle of 15.0° above the horizontal and a slope inclined at 50.0°. The equations used include the tangent function for slope (tan50° = Yf/Xf) and kinematic equations for vertical and horizontal motion. The user initially miscalculated the vertical displacement (Yf) by not using the correct sign, which led to an incorrect time of flight. The correct time of flight is determined to be 2.88 seconds, which is essential for calculating the landing distance accurately.

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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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