# Energy Slope Question

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1. Oct 9, 2016

### BrainMan

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!

2. Oct 9, 2016

### Staff: Mentor

I think you might want to re-evaluate that. If her direction started out at -60° with respect to the horizontal, adding 90° to that yields...?

3. Oct 9, 2016

### PeroK

Does that look to you like 60 degrees in the diagram?

4. Oct 9, 2016

### BrainMan

OK I get 30 degrees. Following the same process with 30 degrees I get the right answer. Thanks!

5. Oct 9, 2016

### Staff: Mentor

Happy to help!