- #1

OTSEngineer

- 10

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- Homework Statement
- A skier starts from rest at the top of a hill. The skier coasts down the hill and up a second hill, as the drawing illustrates. The crest of the second hill is circular, with a radius of 32.8m. Neglect friction and air resistance. What must be the height h of the first hill so that the skier just loses contact with the snow at the crest of the second hill.

- Relevant Equations
- r=32.8m

Solve for height h

See a picture of the question above.

My thoughts are:

- dp(y)/dy is negative such that when going up the slope, the momentum in the y direction is equal to 0 just as the skier reaches the top of the circular section.
- Given that there is no friction on the slopes, the energy of the skier (potential+kinetic) is the same across the dashed line for all time.
- The ground does not have a force acting on the skier when it reaches the top, because the skier had just enough momentum to reach the peak without interacting with the snow.

I'm thinking that the potential energy gained by starting at height h is equal to the kinetic energies at the two points where the slopes and the dashed line intersect. As such, I could say that the kinetic energy at the first intersection point is equal to the kinetic energy at the second; 1/2*m*(mag(v))^2 at first intersection would then be equal to 1/2*m*(vy)^2 at the second. And, that the kinetic energy at the first point is equal to the potential energy at the starting point. I think that the angle between the velocity vectors at the first intersection point would determine height h, however, I'm not certain how to find that value with radius r. I think that the length r is the length of the hypotenuse of the two velocity vectors at the first intersection point, but I need 1 extra metric about the triangle (angle or the length of one side) to calculate h. Maybe I'm missing something? Or maybe my assumptions are incorrect?

Thanks,

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