Verified answer

This is no easy problem.

see http://ceprofs.tamu.edu/rbruner/vertical/sld008.ht...

The general equation is y(x) = a x^2 + g1 x + c, let c = 0 (initial elevation)

a is a function of grade in, g1, grade out, g2 (which is negative) and the length of the curve.

Since grade in is positive and equals grade out we can solve for a in terms of g1, since a = r/2 and since r = (g2 - g1)/L = - 2 g1/L. Since L = 40, the r = - g1/20. Hence a = - g1/40

Substituting then we get Y (x) = - g1/40 x^2 + g1 x for our formula.

Follows that 8 = - g1/40 * 5^2 + g1 * 5, and solving for g1 we get:

g1 = 1.828 and a = -g1/40 = -0.0457 for our constants.

Finally, for Y (12) = 15.35 m TA DA lol