Position vector for free fall along an inclined plane

Question

I understand how to make a force diagram for an inclined plane.

Assuming I have an inclined plane in the form of a right-angled triangle with both cathetes $k$, the angle of the inclined plane is $\tan (k / k) = \tan 1 = 45\deg$. On top I have an object that glides frictionless down the inclined plane (on the hypotenuse). How can I set up the position vector $\mathbf{r}$ for $x$ and $y$ for the uniformly accelerated movement?

If the object were to slide down vertically, $\mathbf{r} = (0, k-\frac{1}{2}gt ^ 2)$.

Answer 1

This is not the most convenient coordinate system but any system should do. You can start with what you know: the displacement along the inclined as a function of time. So just take a point somewhere on the inclined plane. This will be at a distance d from the top of the inclined. The distance d is easy to find from the equation of motion with acceleration $g \sin( \theta) $. Then is just a straightforward geometry problem to find the x and y coordinates of this point. You draw two lines from this point, each parallel to one of the axes. Two smaller triangles will form. Use trigonometry to find k-x and k-y. These will be sides of the small triangles.