Kinetic energy for a hydrostatic fluid versus a nonhydrostatic fluid I am reading in my textbook that the kinetic energy of a non-hydrostatic fluid parcel is given by $\boldsymbol{v}^2/2$, where $\boldsymbol{v}$ is the velocity with components $(u,v,w)$, but for a hydrostatic fluid parcel the kinetic energy is defined as $\boldsymbol{u}^2/2$, where $\boldsymbol{u}=(u,v)$. I assume that the hydrostatic equation thus makes vertical velocity negligible comapared to horizontal velocities - but I cannot wrap my head around why this occurs, or how the hydrostacy of a fluid reduces vertical motions.
My one guess is that a fluid balanced in the vertical does not deviate much from this balance, thus reducing vertical motions, but I'm not sure about this, since there is no  vertical velocity term $w$ in the hydrostatic balance. 
 A: Hydrostatics is a branch of fluid dynamics dedicated to fluids at rest. Furthermore the property as described by you would be a property of the flow not the fluid itself. In simple shear flow (Couette flow) for example you may assume that there will be no vertical motion throughout the entire domain because there can't be a vertical velocity component at the top as well as at the bottom boundary. As you seem to come from the field of geoscience and meteorology I will use that as an example.
Atmospheric transport
In atmospheric transport there are huge differences in latitudinal, meridional and interhemispheric transport rates as they have different causes, mechanisms and different boundary conditions.
Boundary conditions: The bottom of the earth acts approximately like a solid no-slip wall where all the velocities approach zero. The top is not constrained in latitudinal and meridional (horizontal) direction but in vertical direction the velocity can be assumed zero due to gravity (the fluid can't really leave the field of gravity).


*

*The latitudinal motion (around $10 \frac{m}{s}$) is caused by the rotation of the earth and the inertia of the air as well as local differences in atmospheric pressure.

*The meridional flow (around $1 \frac{m}{s}$) is mainly driven by differences in atmospheric pressure and is thus a lot lower. 

*The vertical motion ($0.001-0.01 \frac{m}{s}$ while it can be higher locally due to buoyancy) is mainly a reaction to the other two motions due to continuity reasons (convergence and divergence in horizontal direction must be equilibrated by vertical motion).
You can see the vertical motion can be most likely regarded as negligible as it is several orders of magnitude smaller than the other two components.
A note on simplifications
In the end such assumptions about the flow are merely acts of desperation: Realistic flows are impossible to solve analytically and time-resolved 3D simulations - if you neglect RANS simulations - are computationally too expensive to solve numerically for most applications as well. You will try to reduce complexity by making assumptions about the nature of the fluid and flow (incompressible flow or fluid, neglect certain terms as they have a small order of magnitude, reducing dimensions etc.). Nonetheless this will be far from true for most applications. In particular if you have aerodynamic flows they will be highly turbulent and a symmetry can only be imposed for time-averages but not for velocity fluctuations. 
For example in car aerodynamics - if obtaining estimates by hand analytically - you will assume that your car geometry is symmetric ("indefinitely long") and look at a 2D slice. You will further apply simplified concepts such as stationary inviscid (Euler equations) flow. The viscous effects are negligible small for $Re \to \infty$ far from the boundaries but actually not close to the walls, a car won't be sufficiently modelled by a 2D model as you will have things like a horseshoe vortex and the flow won't be stationary either. Nonetheless you might be able to get good estimates.
