Calculating Ball's Range on Incline 15 Degrees

[P.O.L.A.R]

Homework Statement

A ball is kicked with initial speed 20 m/s and initial angle 40 degrees up an incline of angle 15 degrees. Assume that the ball leaves the ground a the base of the incline at $$x_{0}$$=0 and $$y_{0}$$=0. How far up the incline does the ball initially land (not how far horizontally or vertically but how far along the incline)?

Homework Equations

Range: $$R=(v^{2}_{0}/g)sin2\Theta_{0}$$
y-$$y_{0}=(tan(\Theta_{0})(x-x_{0})-g(x-x_{0})^{2}/2(v_{0}cos\Theta_{0})^{2}$$

The Attempt at a Solution

Well what I tried was subtracting 15 from 40 and came up with 25 plugged it into the Range formula and went from got 31.3m. I think I am missing the the 15 degree incline and was wandering if I just multiplied the range by cos(15)?

I also thought that finding the slope of the 15 degree line then setting it equal to the trajectory formula I could find the point of intersection and do some trig from there. In order to do that I would have to find the slope of the line. I was wandering if [sin(15)/cos(15)]x would be the slope of the 15 degree incline?

Not sure which method works the first one seems like it could work but I was wondering if gravity changes when the angles are subtracted and if multipling by cos(15) is needed to make up for the incline?

 

[learningphysics]

I recommend working the problem with the x-axis along the direction of the incline... the y-axis perpendicular to the incline... ie what is vx initial... what is vy initial... what is the acceleration in the x direction... what is the accleration in the y direction.

then it is just a straight projectile problem.

 

[P.O.L.A.R]

I recommend working the problem with the x-axis along the direction of the incline... the y-axis perpendicular to the incline... ie what is vx initial... what is vy initial... what is the acceleration in the x direction... what is the accleration in the y direction.

then it is just a straight projectile problem.

Ok that makes total sense thanks