Ok, I'm vexed... I was given the following problem:

The water is shown leaving the nozzle along a 3-4-5 triangle where the horizontal component is 4 and the verticle component is 3 (or 36.1'ish degrees).

So, my question is how do I find velocity given an angle and a curvature? I know radius of curvature ([itex]\rho[/itex]) is:

I have no idea where to go from here. My professor told us the answer was 16.57m/s but I haven't the foggiest clue on how to get there. Any help would be greatly appreciated.

Is a radius of curvature even defined? The water projects in a parabolic arc, not a circular arc. The radius of curvature changes constantly from the very beginning.

What do you mean defined? The initial curvature [itex]\rho=35m[/itex] was given in the problem. There is no functional definition of curvature but I dare say that ties in with the projectile motion equations in some manner. I just don't know how to do it. My text doesn't cover anything like this either. All of the projectile motion questions in the text are presented with straightforward initial conditions.

Yes, the radius of curvature is defined at each point, given by the formula faust9 gave. The radius of curvature is given at the nozzle and the question asks for the radius of curvature at the vertex of the parabola.

JohnDubYa's point that this is a parabola is important: set up a coordinate system so that the nozzle is at (0,0) and and the y axis is parallel to the axis of the parabola. You can write the parabola as y= ax^{2}+ bx. y'= 2ax+ b which, at x= 0, is y'= b. You can find b from the "3- 4- 5" information. y"= 2a and you find a from the curvature at x= 0.

Once you know the equation of the parabola you can use the fact y= -(g/2)t^{2}+ v_{y}t and x= v_{x}t to find v_{x} and v_{y[\sub] (the components of velocity at t= 0) to find the initial velocity.}