= Solution
Let $x$ measure distance down the incline and $y$ distance perpendicular to it, with the wall at $y=0$ and the <free surface> at $y=h$. A uniform steady <falling film flow> has $\mathbf u=u(y)\mathbf e_x$. Its <convective acceleration> vanishes because $\mathbf u\cdot\nabla=u\partial_x$ and $\partial_xu=0$. A flat surface exposed to constant atmospheric <pressure> has
$$
p(y)=p_{\rm atm}+\rho g\cos\alpha\,(h-y).
$$
Thus $\partial_xp=0$, and the downslope component of the <Navier-Stokes equations> reduces to $0=g\sin\alpha+(\eta/\rho)u''$. The printed pressure-free equation must be understood as this \b[downslope projection]: its full vector version would incorrectly omit the normal <hydrostatic pressure> gradient that balances $g\cos\alpha$.
The <no-slip boundary condition> gives $u(0)=0$. Neglecting air <shear stress> makes the surface shear-free, so $u'(h)=0$. Two integrations give the <uniform gravity-driven Newtonian film> profile
$$
u'(y)=\frac{\rho g\sin\alpha}{\eta}(h-y),\qquad
u(y)=\frac{\rho g\sin\alpha}{\eta}\left(hy-\frac{y^2}{2}\right).
$$
The <velocity> increases up to the <free surface>, where
$$
\boxed{u_{\max}=\frac{\rho gh^2\sin\alpha}{2\eta}.}
$$
This assumes a uniform laminar <Newtonian fluid> film, a stationary wall and negligible surface shear; a varying thickness or externally imposed downslope <pressure gradient> would alter the balance.
\Image[/past-exam-of-the-astrophysical-fluid-dynamics-course-of-the-university-of-cambridge/2013/paper-1-inclined-film.png]
{title=Uniform inclined film: surface speed maximum and normal hydrostatic pressure}
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