= Solution
{c}
Take $z$ positive away from the wall into the liquid, and write $p_a$ for <atmospheric pressure>. For a constant-<mass density> <Newtonian fluid>, the two-dimensional <Navier-Stokes equation> and <continuity equation> are
$$
\begin{aligned}
u_x+w_z&=0,\\
\rho(u_t+uu_x+wu_z)&=-p_x+\mu(u_{xx}+u_{zz})+\rho g\sin\alpha,\\
\rho(w_t+uw_x+ww_z)&=-p_z+\mu(w_{xx}+w_{zz})-\rho g\cos\alpha.
\end{aligned}
$$
The stationary wall imposes the <no-slip boundary condition> and <no-penetration boundary condition>, $u=w=0$ at $z=0$. For a flat <free surface> with negligible air <shear stress>, the <stress boundary condition> is $u_z=0$, $p=p_a$ at $z=h_0$; the <kinematic boundary condition> is $w=0$ there.
For <steady flow> independent of $x$, these equations give $w=0$, $\mu u_{zz}=-\rho g\sin\alpha$ and $p_z=-\rho g\cos\alpha$. Two <integrations> and the <boundary conditions> give the <uniform gravity-driven Newtonian film>:
$$
\boxed{\bar u=\frac{\rho g\sin\alpha}{2\mu}(2h_0z-z^2),\qquad
\bar w=0,\qquad \bar p=p_a+\rho g\cos\alpha(h_0-z).}
$$
Its surface <velocity> is $U_0=\rho gh_0^2\sin\alpha/(2\mu)$, while its mean <velocity> is $2U_0/3$.
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