Determine the minimimum depth of a water reservoir that will provide a flow rate of 1200 gpm in a 4 inch diameter horizontal pipe that opens to the atmosphere

You ought to consider Bernoulli's equation for this, it is in your text.

You can model it as this:

A drop of water at the top has potential energy m*g*h
This has to equal the kinetic enery of the same drop going out of the horizontal tank
1/2 m v^2

Now veloicty v can be manipulated equal
area*velocity = volume/time, and you know the volume/time as 1200g/min. Change that to m^3/sec.
set the two energies equal.
1/2 mv^2=mgh
h= 1/2 v^2/g
put in for v the velocity (convert 1200gal/min to inch^3/sec, divide by the area of a 4 inch pipe. Then solve for h.

im sorry but i am completly lost.
i have h= 1/2 v^2/g
i know h is height v is velocity and g is gravity but how do i find g?
another thing what is the easiest way to convert gal/min to inch/sec?

gal convert to inches cubed.
min convert to sec

To find the value of gravity (g), you can use the standard value of acceleration due to gravity, which is approximately 9.81 m/s² or 32.2 ft/s². If you're working in different units, you may need to convert it accordingly.

To convert gallons per minute (gal/min) to cubic inches per second (in³/s), you can use the following conversion factors:

1 gallon = 231 cubic inches
1 minute = 60 seconds

So, to convert gal/min to in³/s, you can multiply the value in gal/min by (231 in³/gal) / (60 s/min). This will give you the conversion factor.

Once you have the conversion factor, you can calculate the value of h in the equation h = 1/2 * v² / g, where v is the velocity and g is the acceleration due to gravity.