1. The problem statement, all variables and given/known data It's been a great day of new, frictionless snow. Julie starts at the top of the 60∘ slope shown in the figure (Figure 1) . At the bottom, a circular arc carries her through a 90∘ turn, and she then launches off a 3.0-m-high ramp. How far horizontally is her touchdown point from the end of the ramp? Express your answer to two significant figures and include the appropriate units. 2. Relevant equations (1/2)mv2i + mgyi = (1/2)mv2f + mgyf 3. The attempt at a solution Solving for V at the end of the ramp I got V = sqrt(2g(y1 - y2)) So the velocity as she is going off the ramp should be Vf = sqrt(2g(25-3)) = 20.8 m/s Because of the 90 degree arc length the angle she is going off the ramp should also be 60 degrees. So y = y0 + v0t - (1/2)gt2 0 = 3 + 20.8 sin (60) - 4.9t2 0 = 3 + 18.01t - 4.9t2 t = 3.83 plugging into the x equation for distance x = vt = v0cos(θ)t = 20.8 * cos(60) * 3.83 = 40 m I'm not sure why this isn't right. Any help is greatly appreciated!