Reading up on Lagrangian mechanics, it's fascinating. Entirely different view, one single rule, a complete alternative to Newton's laws. But how do you actually find the path of least action?
Let's say I were trying to find the path taken by a projectile launched by a cannon, image courtesy of Dr. Thomas Gibson, Texas Tech University:
Let's say the above cannon launches a projectile off a cliff at $0^\circ$ off the horizontal. The mass doesn't actually affect the path disregarding air resistance, but it's needed for Lagrangian mechanics so let's say it's a $10\textrm{kg}$ projectile. The projectile is fired from $10\textrm{m}$ off the ground, and the velocity imparted to the projectile is $10^{\textrm{m}}/_{\textrm{s}}$ entirely in the positive $x$ direction.
Newtonian
Running through this Newtonian-style, if we wanted to find the location of the projectile at any given point in its path, we know that the horizontal component of its position can be found using the initial velocity, and the vertical component can be found using acceleration due to gravity, so:
$$\vec{s}(t) = \left \langle {v_xt}, {\frac{gt^2}{2}} \right \rangle$$
Lagrangian
For the Lagrangian approach, we know that the kinetic energy of the projectile at the start is $\frac{mv^2}{2}$ and ignoring air resistance the potential energy of the projectile at the start is $mgh$. So, within the context of a given path,
$${\mathcal {S}}(L)=\int_{t_i}^{t_f}{\left[ \frac{mv_x^2}{2} - mgh \right]} \,dt$$
And the path of least action $L_{LA}$ from all possible paths $L$ is defined as:
$$\{L_{LA} \in L \mid {\mathcal {S}}(L_{LA}) = \min_{L_k \in L}{\mathcal {S}}(L_k)\}$$
So I've got my definition down, but how do I actually find the curve followed by the projectile:
$$\vec{s}(t) = \left \langle {?}, {?} \right \rangle$$
Example
Let's say I wanted to find the position of the projectile at the middle of its path, by time.
In the Newtonian model, I know that the projectile's path will end upon hitting the ground, so the total time in the air can be expressed as $t_f = \sqrt\frac{2d}{g}$ so at half of $t_f$ then, $t_{mid} = \sqrt\frac{d}{2g}$.
Substitute $t_{mid}$ in for $t$ and we have the position mid-path by time:
$$\vec{s}(t_{mid}) = \left \langle {v_x\sqrt\frac{d}{2g}}, {\frac{d}{4}} \right \rangle$$
But I can't quite figure out how to arrive at the same result from the Lagrangian form. How would I use the action to actually find details of the favored path?
$${\mathcal {S}}(L)=\int_{t_i}^{t_f}{\left[ \frac{mv_x^2}{2} - mgh \right]} \,dt \quad\quad \Longrightarrow \quad\quad \vec{s}(t_{mid}) = \left \langle {v_x\sqrt\frac{d}{2g}}, {\frac{d}{4}} \right \rangle$$