Viscous Fluids at a Slope (Navier-Stokes) An in-compressible viscous fluid flows down a flat slope of angle θ to the
horizontal under the force of gravity, with g the acceleration due to gravity.
What are the boundary conditions for the fluid at the point of contact with
the slope and at the free surface?
What I know: $u=0$ at $y=0$ and $\frac{du}{dy}=0$ at y=d. Then $\vec {g}=[g sin(a),-gcos(a),0]$
Using orthogonal coordinates with the x-axis pointing down the slope and
the y-axis perpendicular to the slope, find a solution to the Navier-Stokes
equation for a flow of depth d down the slope under the assumptions that
the flow is steady and uniform in the x-direction, including an expression
for the pressure.
This is how i interpreted the question in a figure:

A river descends by 100m over a distance of 100km. Given that the dynamic viscosity of water is approximately $$\mu=10^{−3}kgm^{−1}s^{−1}$$ 
estimate the predicted speed of the river using your own estimates for any other
parameters involved.
Is it unrealistic? If so give possible reasons for the lack of realism.
What I know: is the Navier-Stokes Equation is... $$\dfrac{d\vec{v}}{d t}+ \vec{v} .\nabla \vec{v} = \vec{F} -  \dfrac{1}{\rho} \nabla p + \nu \nabla^2 \vec{v}$$
Now I'm a bit unsure but I think due to the fluid being in-compressible we now have the equation: $$\dfrac{d\vec{v}}{d t}+ \vec{v} .\nabla \vec{v} = \vec{F} -  \dfrac{1}{\rho} \nabla p$$
Really stuck from here, on this question don't really know how to approach it as I'm fairly new to fluid dynamics any help would be greatly appreciated.
 A: You are conflating incompressible with inviscid.  Here we have flow of an incompressible viscous fluid and the term $\nu \nabla^2 \mathbf{v}$ may not be neglected.
You are asked to find "a solution" under the assumptions that the flow is steady and uniform in the x-direction. This implies fully-developed unidirectional flow where the only non-vanishing component of velocity is $u$ (i.e., the component in the x-direction).
For an incompressible fluid, the equation of continuity gives
$$\nabla \cdot \mathbf{v} = \frac{\partial u}{\partial x} = 0$$
Thus, $u$ is a function only of the y-coordinate and the Navier-Stokes equations reduce to
$$0 = -\frac{1}{\rho}\frac{\partial p}{\partial x} +g \sin \alpha + \nu \frac{d^2u}{dy^2}, \\0 = -\frac{1}{\rho}\frac{\partial p}{\partial y} -g \cos \alpha $$
Since the flow is gravity-driven, we can neglect the x-component of the pressure pressure gradient $\frac{\partial p}{\partial x}$ in the first equation and find $u$ by applying the two boundary conditions to the solution of the second-order differential equation
$$\nu \frac{d^2 u}{dy^2} = -g \sin \alpha$$
The second equation allows for solution of the  pressure as a function of the y-coordinate.
See if you can finish now.
