= Solution
Put $m_0=M_0/\rho_0$ and $f_0=F_0/\rho_0>0$ under the literal mass convention. Their dimensions are $[m_0]=L^4T^{-2}$ and $[f_0]=L^4T^{-3}$. <Dimensional analysis> gives
$$
\boxed{L_J=\frac{m_0^{3/4}}{f_0^{1/2}}
=\frac{M_0^{3/4}}{\rho_0^{1/4}F_0^{1/2}}}.
$$
The <jet length> compares source <momentum> with <buoyancy>. The region near the source behaves as a <turbulent round jet>; sufficiently far away, the <forced plume> behaves like a <pure plume>. With the constant <entrainment coefficient> retained explicitly, substantial turning occurs over a length of order $\ell_J=L_J/\sqrt E$, where $E=2\alpha\sqrt\pi$. Statements involving $s\ll L_J$ or $s\gg L_J$ normally hold this dimensionless coefficient fixed.
Near the point source, $m\sim m_0$ and the <Batchelor entrainment hypothesis> gives $q\sim E\sqrt{m_0}s$. Therefore
$$
m_z=m_0\sin\theta_0+\frac{Ef_0}{2\sqrt{m_0}}s^2+O(s^4),\qquad
\theta=\theta_0+\frac{Ef_0\cos\theta_0}{2m_0^{3/2}}s^2+O(s^4).
$$
Integrating the centreline tangent gives
$$
\begin{aligned}
x(s)&=s\cos\theta_0-\frac{Ef_0\sin\theta_0\cos\theta_0}{6m_0^{3/2}}s^3+O(s^5),\\
z(s)&=s\sin\theta_0+\frac{Ef_0\cos^2\theta_0}{6m_0^{3/2}}s^3+O(s^5).
\end{aligned}
$$
\b[To leading order the centreline is a straight <turbulent round jet> at its source inclination.] For the horizontal source, the first curvature is the <horizontal forced-plume trajectory>
$$
\boxed{z\sim\frac{Ef_0}{6m_0^{3/2}}x^3
=\frac{\alpha\sqrt\pi}{3L_J^2}x^3}.
$$
For a vertical source, $\cos\theta_0=0$, the trajectory stays vertical rather than developing this cubic transverse displacement. If $F_0=0$, there is no finite buoyancy-induced <jet length> and the <turbulent round jet> remains straight.
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