How can the range of a projectile fired at an angle up a hill be calculated?

[cameo_demon]

the following problem is from the 5th edition of Thorton and Marion's "Classical Dynamics"
ch.2 problem 14 p.92

Homework Statement
A projectile is fired with initial speed v_0 at an elevation angle of alpha up a hill of slope beta (alpha > beta).

(a) how far up the hill will the projectile land?
(b) at what angle alpha will the range be a maximum?
(c) what is the maximum angle?

The attempt at a solution
apparently this has been a stumper in former classical mechanics classes, but here was as far as i got:

i broke down the components of the forces into x and y
x-component:
a_x=0 integrating -->
v_x = v_0 cos(beta) integrating -->
x = v_0 t cos(beta)

a_y = -g integrating -->
v_y = -gt + v_0 sin (alpha - beta) integrating -->
y = ( -gt^2 / 2 ) + v_0 sin (alpha - beta)


the answer in the back of the book for part (a) is:
d = (2 v_0^2 cos(alpha) sin(alpha-beta) ) / (g cos^2(beta))

any idea how to get from the components to the answer?

 

[Hootenanny]

Hint: The distance is the magnitude of the position vector.

 

[msd213]

Hint: The distance is the magnitude of the position vector.

I too am having trouble with this problem. In the book, they use Taylor expansions in order to find the flight time T and put in back into the range equation. However, they have the luxury of letting y = 0 because the trajectory is a simple parabola. We don't have the same situation here. y doesn't equal zero, y = beta*range.

 

[msd213]

I have no idea how we're supposed to get things like cos($$\beta$$)