This is a two part physics/math question. I am interested in the math
formulas to compute the answers. Given: 12 foot tall open top water
tank; Two holes in the side of the tank at the bottom edge, one is
.250" and other is 1.128"; water level is held constant at 12 feet;
tank is on a stand so the ground won't interfer with the out streams.
PART 1: How far out will the warter squirt horizontally? PART 2:
Enclose top of tank. Hold 100 psi on the water column. Water level
held constant at 12 feet. How far now?

Hi Rob
             It was some time

  The water is in free fall, meaning" it?s trajectory is same is a
ball thrown horizontally would be


here is  Free fall   = general intro
       http://www.learner.org/exhibits/parkphysics/freefall.html   

and here is graphic illustration

   http://www.phys.virginia.edu/classes/109N/more_stuff/Applets/newt/newtmtn.html

and here is how you calculate the distance.  

Key word for understanding the formula is 'decomposition'   which means:

 The  trajectory (which is parabola) is composed of     vertical  
Uniformly Accelerated Motion

    http://online.cctt.org/physicslab/content/Phy1/lessonnotes/acceleratedmotion/lessonacceleration.asp


   y=s = v*  t + ½ a * t^ 2 , where a in our case is g (gravity acc 9.81 m/ s * s)

and horizontal uniform motion ( same with a=0) i.e.

 x= vh * t


So, to get distance of impact

  You take  y = distance of the hole to the ground, calculate t 
(initial velocity v was zero)

 having the t, you calculate x using vh= initial horizontal velocity. 

  What is vh depends on the pressure at the opening, which is any
added pressure ( 100 psi) and
 pressure caused by the water column = weight of water over area of the column.

 How does it depend on the pressure p?
 
Conservation of energy  will help here:  Imagine hole closed. Remove
one cc (cubic centimeter) of water from the bottom.
The water column level will go down a bit. That amount of potential
energy ( m g h ) is converted into kinetic energy
of    that 1 cc, when it squirts out ( 1/2 m v *v ). This v is you vh.

We are neglecting viscosity here. For water it is OK; obviously effect
would be different with jar of honey.

Hedgie

How high is the stand?

I don't care, say one foot. Thanks! Robert Williams

Without a pressure head, and neglecting friction, the water will
squirt out just fast enough so that, if the stream were directed
straight up, it would reach the level of the surface of the water.  So
it comes out of the holes with a velocity v = sqrt(2gh) where g is the
acceleration of gravity, 32.2 ft/sec^2, and h is the height of the
water.  I'm using English units rather than metric, so some
conversions are required.  The effective height added to the column of
water by pressurizing the air above it is H = p/gd, where p is the
pressure and d is the density of the water.  Here come those
conversions: p is 100 psi, but because our unit of length is the foot,
we need pressure in pounds per square foot.  Since there are 144 sq
inches in a sq foot, the pressure is 14400.  d is the mass density of
water, 1000 kg/m^3, which converts to 1.94 slug/ft^3 (slug is the
English unit of mass and is equal to [lb * sec^2 / ft]).

So:

H = p/gd = (14400 lb/ft^2) / (32.2 ft/sec^2 * 1.94 slug/ft^3) = 230 ft

v1 = sqrt(2gh) = sqrt(2 * 32.2 ft/sec^2 * 12 ft) = 27.8 ft/sec

v2 = sqrt(2g(h + H)) = sqrt(2 * 32.2 ft/sec^2 * (12 ft + 230 ft)) = 125 ft/sec

Now, from distance = .5 * a * t^2, the water will fall one foot in 0.25 sec.

So in the first case the water will go 27.8 * .25 = 6.93 ft horizontally.

In the second case the water will go 125 * .25 = 31.2 ft horizontally.

Note that the diameter of the hole does not effect the outcome, as
long as it is big enough so viscosity is not important.

You could also use Bernoulli's equation:

.5 * v^2 + gz + p/d = constant    (z is the vertical coordinate)