= Solution
The required <Hamilton-Jacobi equation> is
$$
\boxed{S_t+\frac1{2m}(S_x^2+S_y^2+S_z^2)+mgz=0.}
$$
Because $x,y$ and time are cyclic in the <Hamiltonian>, use separation $S=ax+by-Et+W(z)$. The remaining equation is
$$
(W')^2=2m(E-mgz)-a^2-b^2\equiv Q(z).
$$
For $g\ne0$, each sign branch has the separated complete integral
$$
\boxed{S=ax+by-Et-\frac{s}{3m^2g}[2m(E-mgz)-a^2-b^2]^{3/2},\qquad s=\pm1.}
$$
It is real in the classically allowed region $Q\ge0$, and $p_z=s\sqrt Q$. At a turning point one matches the ascending and descending branches. If $g=0$, the integral is simply $W=p_zz$ with constant $p_z$.
A convenient equivalent complete integral, expressed using initial-momentum constants $\boldsymbol\alpha$, is globally polynomial in time:
$$
\boxed{\begin{aligned}
S={}&\alpha_xx+\alpha_yy+(\alpha_z-mgt)z
-\frac{\alpha_x^2+\alpha_y^2+\alpha_z^2}{2m}t\\
&+\frac{\alpha_zg}{2}t^2-\frac{mg^2}{6}t^3.
\end{aligned}}
$$
Substitution verifies that its time derivative cancels the kinetic term and $mgz$. It avoids a square-root branch change when extracting the complete <trajectory>.
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