Sorry for the maybe silly question, but I'm a math student and I'm not able to find the answer to this question from the physics point of view.
Let us suppose that a ball is set in a non equilibrium state in a mountain, with $0$ initial velocity. Let us treat the ball as a point and let $M=\{(x,y,z)|f(x,y)=z\} \subset \mathbb{R}^3$ be the profile of the mountain. Physically I think that the trajectory followed by the ball will be as follows: let $\gamma(t)=(x(t),y(t),f(x(t),y(t))$ be the trajectory curve such that at $t=0$ the point $\gamma(0)$ is where we set the ball. Since the gravity is pointing down in the $z$ direction we have that $\gamma(t)$ satisfies the following differential equation $$\gamma^{"}(t)=\Pi_{\large{T_{\gamma(t)}M}}(-g(0,0,1))$$ where $\Pi:\mathbb{R}^3 \rightarrow T_{\gamma(t)}M$ is the linear projection and $g$ is the classical gravitational constant. But clearly there is another "canonical" trajectory attached to $M$ and to the point $\gamma(0)$, i.e. the unique geodesic $\Gamma \subset M$ starting at $\gamma(0)$. Clearly by definition the geodesic $\Gamma$ is parametrized by arclength, while the physical trajectory $\gamma(t)$ certainly not.
However is it true that in general $\gamma(t)$ and $\Gamma$ have the same support as subsets of $M$? In the negative case is possible to find a surface $S$ embedded in $\mathbb{R}^3$ such that that geodesics of $S$ are physical trajectory of non equilibrium points of $M$?
