= Solution
For steady <inviscid flow> at atmospheric <pressure>, <Bernoulli's equation> along a freely falling <streamline> gives $u^2/2+gz=u_0^2/2+gH$. The downward speed is therefore
$$
\boxed{u=u_0\left[1+\frac{2g}{u_0^2}(H-z)\right]^{1/2}}.
$$
For the incompressible jet, <mass conservation> fixes its <volume flux>: $\pi R^2u=\pi a^2u_0$. Consequently
$$
\boxed{R=a\left[1+\frac{2g}{u_0^2}(H-z)\right]^{-1/4}}.
$$
Far from impact, the <inviscid falling jet and radial impact film> is thin and nearly horizontal. Let $v$ be its radial speed and $h$ its thickness. The same <volume flux> now crosses a cylindrical surface of area $2\pi rh$, giving $v=a^2u_0/(2rh)$. At the <free surface>, atmospheric <pressure> and <Bernoulli's equation> give $v^2=u_0^2+2g(H-h)$. Eliminate $v$ to obtain
$$
\boxed{\frac{a^4}{4r^2h^2}=1+\frac{2g}{u_0^2}(H-h)}.
$$
This description assumes a steady incompressible jet, negligible surface tension, and negligible dissipation between the tube and the thin far-field film. It is a free-surface relation, not an assertion that <pressure> is atmospheric throughout the film or that the impact region itself is thin.
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