= Solution
For steady nonzero flux, $Aw=Q$, so $A=Q/w$. Substitution in the <slender viscous thread> force balance gives
$$
\left(\frac{w_z}{w}\right)_z=-\frac{U}{w},\qquad U=\frac{(\rho-\rho_a)g}{3\mu}.
$$
If $f=w_z/w$, then
$$
\frac d{dz}\left(\tfrac12f^2-\frac Uw\right)
=f\left(-\frac Uw\right)+\frac{Uw_z}{w^2}=0.
$$
\b[Hence $\boxed{\tfrac12(w_z/w)^2=U/w+c}$.] This argument does not divide by $w_z$, so it also applies at turning points of the axial velocity. The symbol $U$ here has dimensions inverse length-time; it is not itself a speed.
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