\subsection{Projectile Motion via Hamilton-Jacobi}

This subsection solves the Hamilton--Jacobi equation for projectile motion in a uniform gravitational field, demonstrating that Jacobi's theorem reproduces the standard parabolic trajectory without ever integrating Newton's second-order differential equations. The kinematics approach (Unit 1, Section m1--5) solves two decoupled ODEs for $x(t)$ and $y(t)$ separately. Here, a single first-order PDE separates into the same two independent problems because the cyclic coordinate $x$ forces the horizontal--vertical split by the structure of the formalism. The energy budget, cross-referencing Unit 3, emerges naturally from the separation constants rather than from the work--energy theorem.

\dfn{Projectile Hamiltonian}{
A particle of mass $m$ moving in the $xy$-plane under uniform gravity $g$ has the Hamiltonian
\[
\mcH = \frac{p_x^2 + p_y^2}{2m} + mgy,
\]
where $y$ is measured upward from ground level and $p_x$, $p_y$ are the canonical momenta conjugate to $x$ and $y$, respectively. The corresponding Hamilton--Jacobi equation $\mcH + \pdv{\mcS}{t} = 0$ reads
\[
\frac{1}{2m}\left[\left(\pdv{\mcS}{x}\right)^2 + \left(\pdv{\mcS}{y}\right)^2\right] + mgy + \pdv{\mcS}{t} = 0.
\]}

\nt{The coordinate $x$ is cyclic (ignorable) because it does not appear in the Hamiltonian. Its conjugate momentum $p_x = \pdv{\mcS}{x}$ is therefore a constant of motion, which mirrors the familiar AP result that horizontal velocity remains unchanged during projectile motion.}

\nt{Energy budget: the total energy $E$ splits into a horizontal part $E_x = \alpha_x^2/(2m) = \tfrac{1}{2}mv_{0x}^2$, which is constant because $x$ is cyclic, and a vertical part $E_y = \tfrac{1}{2}mv_y^2 + mgy$. The separation constant $E_x$ is the horizontal kinetic energy, carrying no potential contribution. The vertical energy $E_y$ accounts for both the vertical kinetic and gravitational potential energy, so $E_y$ is conserved within the vertical subsystem. The total energy is $E = E_x + E_y = \tfrac{1}{2}m(v_{0x}^2 + v_{0y}^2)$, matching the initial kinetic energy at ground level. This energy partition is equivalent to the mechanical-energy bookkeeping used in Unit 3.}

\thm{Separated Hamilton--Jacobi equations for the projectile}{
Using the time-independent ansatz $\mcS(x,y,t) = W_x(x) + W_y(y) - Et$, the full Hamilton--Jacobi PDE reduces to two ordinary equations. Because $x$ is cyclic, $\der{W_x}{x} = \alpha_x$ (constant). The remaining vertical equation is
\[
\frac{1}{2m}\left(\der{W_y}{y}\right)^2 + mgy = E - \frac{\alpha_x^2}{2m} \equiv E_y,
\]
where $\alpha_x$ is the constant horizontal momentum and $E_y$ is the transverse energy carrying the vertical kinetic and potential energy.}

\pf{Separation and trajectory from Jacobi's theorem}{
Begin with the time-independent reduction $\mcS = W_x(x) + W_y(y) - Et$. Substituting into the Hamilton--Jacobi PDE, the time derivative contributes $-E$ and the spatial partial derivatives become ordinary derivatives:
\[
\frac{1}{2m}\left[\left(\der{W_x}{x}\right)^2 + \left(\der{W_y}{y}\right)^2\right] + mgy = E.
\]
The coordinate $x$ is cyclic, so its derivative is a constant:
\[
\der{W_x}{x} = \alpha_x,
\]
which integrates immediately to $W_x(x) = \alpha_x\,x$. Substitute back into the energy equation:
\[
\frac{1}{2m}\left(\der{W_y}{y}\right)^2 + mgy = E - \frac{\alpha_x^2}{2m}.
\]
Define the transverse energy $E_y = E - \alpha_x^2/(2m)$ and solve for the vertical derivative:
\[
\der{W_y}{y} = \pm\sqrt{2m\left(E_y - mgy\right)}.
\]
Integrate with respect to $y$. Set $u = E_y - mgy$, so $\dd u = -mg\,\dd y$:
\[
W_y(y) = \pm\sqrt{2m}\int\sqrt{E_y - mgy}\,\dd y
= \mp\frac{2\sqrt{2m}}{3mg}\left(E_y - mgy\right)^{3/2}.
\]
Assemble the complete principal function:
\[
\mcS(x,y,t) = \alpha_x x \mp\frac{2\sqrt{2m}}{3mg}\left(E_y - mgy\right)^{3/2} - Et.
\]
Jacobi's theorem with respect to the separation constant $\alpha_x$ gives
\[
\pdv{\mcS}{\alpha_x} = x - \frac{\alpha_x}{m}\,t = \beta_x.
\]
Solving for $x(t)$ with $\beta_x = 0$ (launch from the origin):
\[
x(t) = \frac{\alpha_x}{m}\,t = v_{0x}\,t.
\]
To find $y(t)$, apply Jacobi's theorem with respect to $E$. The principal function depends on $E$ both explicitly in the term $-Et$ and implicitly through $E_y(E) = E - \alpha_x^2/(2m)$. The chain rule gives
\[
\pdv{\mcS}{E}
= \pdv{\mcS}{E_y}\,\pdv{E_y}{E} - t.
\]
Since $\pdv{E_y}{E} = 1$, substituting the $W_y$ term yields
\[
\pdv{\mcS}{E}
= \mp\frac{2\sqrt{2m}}{3mg}\cdot\frac{3}{2}\left(E_y - mgy\right)^{1/2} - t = \beta_E,
\]
which simplifies to
\[
\mp\frac{\sqrt{2m}}{mg}\sqrt{E_y - mgy} - t = \beta_E.
\]
To solve for $y(t)$, choose the sign consistent with an upward launch: $p_y = \der{W_y}{y} > 0$ at $t = 0$ selects the upper square root for $\der{W_y}{y}$, which gives the negative sign in $\mcS$. Rearrange:
\[
\frac{\sqrt{2m}}{mg}\sqrt{E_y - mgy} - t = \beta_E.
\]
At $t = 0$ with $y = 0$ and $v_{0y} = \sqrt{2E_y/m}$, the integration constant is fixed:
\[
\frac{\sqrt{2m}}{mg}\sqrt{E_y} = -\beta_E
\qquad\Longrightarrow\qquad
\beta_E = -\frac{\sqrt{2E_y/m}}{g} = -\frac{v_{0y}}{g}.
\]
Substitute $\beta_E$ back and isolate the radical:
\[
\frac{\sqrt{2m}}{mg}\sqrt{E_y - mgy} = \frac{v_{0y}}{g} - t.
\]
Multiply by $mg/\sqrt{2m}$ and square both sides:
\[
E_y - mgy = \frac{m}{2}\left(v_{0y} - gt\right)^2.
\]
Since $2E_y/m = v_{0y}^2$, divide through by $m$ and expand the right-hand side:
\[
\frac{1}{2}v_{0y}^2 - gy = \frac{1}{2}\left(v_{0y}^2 - 2v_{0y}gt + g^2t^2\right).
\]
The term $\tfrac{1}{2}v_{0y}^2$ cancels, leaving $gy = v_{0y}gt - \tfrac{1}{2}g^2t^2$, so
\[
y(t) = v_{0y}\,t - \frac{1}{2}\,g\,t^2.
\]
The two equations combine into the parabolic trajectory $y = (v_{0y}/v_{0x})\,x - \tfrac{g}{2v_{0x}^2}\,x^2$.}

\nt{Comparison to AP kinematics}{The Hamilton--Jacobi equations $x(t) = v_{0x}t$ and $y(t) = v_{0y}t - \tfrac{1}{2}gt^2$ are identical to the constant-acceleration kinematics of Unit 1 (Section m1--5, free-fall formulas). The separation constant $\alpha_x = mv_{0x}$ is the conserved horizontal momentum, and the transverse energy $E_y = \tfrac{1}{2}mv_{0y}^2$ encodes the initial vertical kinetic energy. Where kinematics integrates $a_x = 0$ and $a_y = -g$ separately into two decoupled ODEs, the Hamilton--Jacobi approach solves one PDE and lets the cyclic coordinate enforce the exact same horizontal--vertical split. The HJ formalism reproduces, from first principles, every projectile-motion result derived elementarily in the AP C syllabus.}

\qs{Projectile launched from the ground}{
A projectile of mass $m = 0.50\,\mathrm{kg}$ is launched from the origin with speed $v_0 = 20\,\mathrm{m/s}$ at angle $\theta_0 = 30.0^\circ$ above the horizontal. Use $g = 9.81\,\mathrm{m/s^2}$.

\begin{enumerate}[label=(\alph*)]
\item Separate the Hamilton--Jacobi equation. Show that $x$ is cyclic and find $\der{W_y}{y}$ in terms of $E_y$ and $y$.
\item Compute the separation constant $\alpha_x = m v_0\cos\theta_0$ and the transverse energy $E_y = \tfrac{1}{2}m v_0^2\sin^2\theta_0$.
\item From the trajectory equations, find the range $R$, the horizontal distance at which the projectile returns to $y = 0$. Verify with $R = v_0^2\sin(2\theta_0)/g$.
\end{enumerate}}

\sol \textbf{Part (a).} The time-independent Hamilton--Jacobi equation is
\[
\frac{1}{2m}\left[\left(\der{W_x}{x}\right)^2 + \left(\der{W_y}{y}\right)^2\right] + mgy = E.
\]
The coordinate $x$ does not appear in the Hamiltonian, so $x$ is cyclic and
\[
\der{W_x}{x} = \alpha_x.
\]
Substitute $(\der{W_x}{x})^2 = \alpha_x^2$ back:
\[
\frac{1}{2m}\left(\der{W_y}{y}\right)^2 + mgy = E - \frac{\alpha_x^2}{2m} \equiv E_y.
\]
Solve for the vertical derivative:
\[
\der{W_y}{y} = \pm\sqrt{2m\left(E_y - mgy\right)}.
\]

\textbf{Part (b).} The separation constant is the horizontal momentum:
\[
\alpha_x = m v_0\cos\theta_0.
\]
Substitute the numerical values:
\[
\alpha_x = (0.50)(20)\cos(30.0^\circ)\,\mathrm{kg\!\cdot\!m/s}.
\]
Using $\cos(30.0^\circ) = \sqrt{3}/2 \approx 0.8660$,
\[
\alpha_x = (0.50)(20)(0.8660)\,\mathrm{kg\!\cdot\!m/s} = 8.66\,\mathrm{kg\!\cdot\!m/s}.
\]
The transverse energy is
\[
E_y = \tfrac{1}{2}\,m\,v_0^2\sin^2\theta_0.
\]
The vertical speed component is
\[
v_{0y} = v_0\sin(30.0^\circ) = (20)(0.500)\,\mathrm{m/s} = 10.0\,\mathrm{m/s}.
\]
Therefore,
\[
E_y = \tfrac{1}{2}(0.50)(10.0)^2\,\mathrm{J} = 25\,\mathrm{J}.
\]

\textbf{Part (c).} The time of flight is found from requiring $y(T) = 0$:
\[
v_{0y}\,T - \frac{1}{2}\,g\,T^2 = 0.
\]
The nonzero root is
\[
T = \frac{2v_{0y}}{g} = \frac{2(10.0)}{9.81}\,\mathrm{s} = 2.04\,\mathrm{s}.
\]
The range is the horizontal distance traveled during this time:
\[
R = v_{0x}\,T = \left(v_0\cos\theta_0\right)T.
\]
The horizontal speed is $v_{0x} = (20)\cos(30.0^\circ)\,\mathrm{m/s} = 17.3\,\mathrm{m/s}$. Hence,
\[
R = (17.3)(2.04)\,\mathrm{m} = 35\,\mathrm{m}.
\]
Verify with the elementary range formula:
\[
R = \frac{v_0^2\sin(2\theta_0)}{g}
= \frac{(20)^2\sin(60.0^\circ)}{9.81}\,\mathrm{m}
= \frac{(400)(0.8660)}{9.81}\,\mathrm{m}
= 35\,\mathrm{m}.
\]
The two results agree to the stated number of significant figures.

Therefore,
\[
\alpha_x = 8.66\,\mathrm{kg\!\cdot\!m/s},
\qquad
E_y = 25\,\mathrm{J},
\qquad
R = 35\,\mathrm{m}.
\]