Jumping Laminar Jets

So a special thanks to my friends who listened to me to help me put together this spreadsheet with all the information about the Jumping Laminar Jets. A special mention to Will for building the spreadsheet for me! For those who aren't fluent in geekspeak what this spreadsheet tells us that it doesn't tell you is how height and far the jumping laminar jets will go based on a number of key numbers.

First column
Flow Rate: This means how much water will flow through an area in a certain amount of time. This is measured in Gallons/Min. It's like how fast you would be going if you were a liquid.

Second column
Flow Rate: Same as above but just converted into different units.

Third column
Angle: This is an important one! Once we build this water fountain we will mount the nozzles at these angles

Fourth and Fifth column
Diameter and Area: The diameter is the diameter of the outlet for the water. Subsequently, the area is the calculated area for the water outlet.

Sixth column
This one is important too. This is the amount of nozzles you can have.

Seventh, Eighth, and Ninth
Velocities. The Seventh column is the total velocity, and the Eighth and Ninth are the component velocities (velocity in the horizontal and vertical directions).

Tenth, Eleventh and Twelfth
Time in the air, Height, and Distance. Pretty self explanatory.

Geekspeak:

It is all based on two principles. First, is flow in must be equal to flow out, Qin = Qout. This obviously simple equation in fact states a lot. From this equation we are able to get the exit velocity of the water because we know what the area is from whence the water is leaving. =)

Simply.

Q=v*A

solving for v

v=Q/A (Eq 1)

so Eq 1 gives us the velocity of the water leaving the nozzle. From there we can treat this like a particle motion problem. Or

h = 1/2*g*t^2 +vy*t+ho (Eq 2)

Since we are going to be putting these nozzles on an angle the velocity that we obtained from Eq 1 isn't the velocity in vo. We need to adjust the velocity for the different angles. So we need to compute the vy (velocity in the y or vertical direction).

vy = vo sin (theta)

knowing that we can calculate the total time the water is in the air. Using Eq 2 we know that the water starts from ground and ends up at the same level (ground). So h=ho = 0. Rearranging Eq 2 and solving for t you get.

t = 2*vy/g Eq 3

Using the answer from Eq 3 and subsituting it back into Eq 2 we can find the height the water will travel.

To find the distance we need one other equation. Distance = Rate * Time

or

D = vx * t

where vx = vo cos(theta) or the velocity in the horizontal direction.

Again, thanks to those who helped me with this!