Finding Touchdown Point Distance on a Slope with a Circular Arc and Ramp Launch

[BrainMan]

Homework Statement

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.

Homework Equations

(1/2)mv²_i + mgy_i = (1/2)mv²_f + mgy_f[/B]

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 V_f = 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!

 

[gneill]

Because of the 90 degree arc length the angle she is going off the ramp should also be 60 degrees.

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...?

 

[PeroK]

Because of the 90 degree arc length the angle she is going off the ramp should also be 60 degrees.

I'm not sure why this isn't right. Any help is greatly appreciated!

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

 

[BrainMan]

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...?

View attachment 107199

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

 

[gneill]

Happy to help!