Step 1: Rewrite $y$ using a kinematic relation, avoiding time altogether.
For vertical motion under constant gravity, $v_y^2 = v_{0y}^2 - 2gy$ (with $v_{0y}=v_0\sin\theta$) lets us write $y$ directly in terms of $v_y$ instead of $t$:
\[ y = \dfrac{(v_0\sin\theta)^2 - v_y^2}{2g} \]
So the Lagrangian $L = \dfrac{1}{2}m(v_x^2+v_y^2) - mgy$ can be written purely in terms of $v_y$ (since $v_x=v_0\cos\theta$ is constant), without writing $y$ or $L$ as an explicit function of $t$.
Step 2: Simplify $L$.
Substituting $y$ from Step 1:
\[ L = \dfrac{1}{2}mv_0^2\cos^2\theta + \dfrac{1}{2}mv_y^2 - \dfrac{m}{2}\big[(v_0\sin\theta)^2 - v_y^2\big] = \dfrac{1}{2}mv_0^2\cos(2\theta) + mv_y^2 \]
Step 3: Integrate by switching the integration variable from $t$ to $v_y$.
Since $v_y = v_0\sin\theta - gt$, $dt = -dv_y/g$; as $t$ runs from $0$ to $T_f$, $v_y$ runs from $v_0\sin\theta$ down to $-v_0\sin\theta$ (the flight is symmetric). The constant piece of $L$ integrates over $T_f = 2v_0\sin\theta/g$:
\[ \int_0^{T_f}\dfrac{1}{2}mv_0^2\cos(2\theta)\,dt = \dfrac{mv_0^3\sin\theta\cos(2\theta)}{g} \]
For the second piece, changing variables:
\[ \int_0^{T_f} mv_y^2\,dt = \dfrac{m}{g}\int_{-v_0\sin\theta}^{v_0\sin\theta} v_y^2\,dv_y = \dfrac{2mv_0^3\sin^3\theta}{3g} \]
Step 4: Add the two pieces and put in $\theta=30^{\circ}$.
\[ S = \dfrac{mv_0^3\sin\theta\cos(2\theta)}{g} + \dfrac{2mv_0^3\sin^3\theta}{3g} \]
At $\theta=30^{\circ}$: $\sin\theta = \dfrac{1}{2}$, $\cos(2\theta)=\cos(60^{\circ})=\dfrac{1}{2}$, $\sin^3\theta = \dfrac{1}{8}$:
\[ S = \dfrac{mv_0^3}{4g} + \dfrac{mv_0^3}{12g} = \dfrac{mv_0^3}{3g} \]
Final Answer:
Comparing with $S=f\times \dfrac{mv_0^3}{g}$ gives $f=\dfrac{1}{3}\approx 0.33$.
\[ \boxed{f = 0.33} \]