> But if you throw it just right, the rock could hypothetically have a trajectory that is perfectly parallel to the mountain slope (in reality, due to things like air resistance, this isn't all that feasible).
Not quite. Ignoring air resistance, any throw you could possibly produce will have a parabolic trajectory, which means it could be parallel with the mountain slope only if the slope itself is parabolic.
Nitpicking here, but the trajectory would be approximately elliptical, not parabolic. A parabolic trajectory would require a uniform gravitational field.
> A parabolic trajectory would require a uniform gravitational field.
A uniform gravitational field is such a good approximation for the situation described that I didn't think of including that proviso. Ballistic trajectory calculations close to the Earth's surface assume constant g.
To clarify for anyone reading, the difference is not constant vs variable magnitude of g, it's constant vs variable direction of g. Gravity is a vector pointing at the (weighted by inverse of distance squared) centre of mass, so once you are operating on a global scale you have to account for the direction of g changing as you move around.
The force always pointing towards a single point is what causes an ellipse to form. If the force always pointed down it would be a parabola, and over short distances on the surface this is a really good approximation.
Only now when re-reading your comment do I realize that you said "any throw you could possibly produce", in which case you are correct. I was thinking that you were talking more about the "fire a stupendously powerful cannon to get an orbital (or closeish) trajectory" situation, in which case the velocities would be high enough to not use a uniform gravitational field. My apologies for the misunderstanding.
Not quite. Ignoring air resistance, any throw you could possibly produce will have a parabolic trajectory, which means it could be parallel with the mountain slope only if the slope itself is parabolic.