You need to know the diameter of the opening. From this you can calculate the speed of the water as it is leaving the Decanter, and therefore its kinetic energy.

That would give an exact answer, but the velocity could also be estimated from the distance the water travels.

If the distance they give is the maximum the stream will travel, we will make the assumption that there is little air drag making the optimum angle for distance 45 degrees.

If the initial speed is V, then now the horizontal and vertical speeds are each V/sqrt(2).

The equation of motion in the vertical direction is:

a[sub]y[/sub]=-g

where a[sub]y[/sub] is the acceleration of the water and g is the acceleration due to gravity. Integrating that gives:

v[sub]y[/sub]=v[sub]y0[/sub]-g*t

where v[sub]y[/sub] is the vertical speed and v[sub]y0[/sub] is the initial vertical speed. When v[sub]y[/sub]=0 it has reached the top of its arc. That will occur at:

t[sub]peak height[/sub]=v[sub]y0[/sub]/g

The total hang time for the water will be twice this value (it takes just as long to come down as it does to go up), or 2*v[sub]y0[/sub]/g.

Now lets look in the horizontal direction. Since there is no acceleration in that direction (remember we are neglecting air resistance because otherwise this becomes very difficult) the equation is quite simple:

x=v[sub]x0[/sub]*t

Here x is horizontal distance traveled and v[sub]x0[/sub] is initial horizontal speed. We know the time it is traveling from above, so we combine those two and get:

x=2*v[sub]x0[/sub]*v[sub]y0[/sub]/g

Remember that both of those speeds are the same and equal to V/sqrt(2). With this substitution we get:

x=V[sup]2[/sup]/g

Rearrange that since we know the distance traveled:

V=sqrt(x*g)

Great. Now we know the speed of the water. Lets combine that with the known flow rater, f, to get a power. Power is energy per unit time:

P=.5*m*V[sup]2[/sup]/t

Lets put that in terms of flow rate where flow rate is volume per unit time and density is mass per unit volume. So, mass per unit time is f*rho where rho is density:

P=.5*f*rho*V[sup]2[/sup]

And now we add the velocity we already calculated:

P=.5*f*rho*x*g

With the max flow rate of 30 gallons per 6-second round (that is still the way they do it, right?), a density for water of 62.42 lb/ft[sup]3[/sup], a distance of 20 feet, and a gravity of 32.2 ft/s[sup]2[/sup] I get:

P=0.391 hp

Assuming I got the conversions right you are not exactly burning things up there. Note that that is on the low end as they probably assume a more horizontal stream of water. You will have major efficiency losses though if you try to convert this to usable power using any medieval technology.

In fact, you would probably do a lot better to just set the thing up high and use the potential energy from the water rather than its kinetic energy as you essentially have as much free potential energy as you want by elevating the thing as high as you want.