What is the velocity of a projectile at a given height.

The derivation I have in my book is :

At a height 'h', v(x) =u cos θ

And v(y) = sqrt(u^2 sin^2θ - 2gh)
therefore Resultant Velocity v = sqrt(v(x) + v(y))
= sqrt((u cosθ)^2 + u^2 sin^2θ - 2gh)
= sqrt(u^2 - 2gh)
Note that this is the velocity that a particle would have a height h if it is projected vertically from ground with u.

NOW WHAT DOES THIS MEAN?? THAT THE VELOCITY DOESN'T DEPEND ON THE ANGLE, JUST THE HEIGHT????? CUZ THEY'VE ELIMINATED θ. BUT AT ONE POINT HERE THE V WILL BECOME 0 IN THE FORMULA BUT THAT DOESN'T HAPPEN IN REALITY UNTIL IT COMES TO REST.

If they've given the wrong thing then please tell the correct formula.

According to what you've provided here, u is the initial velocity of the projectile. When you want to find the resultant velocity at any point, provided a height h, the velocity function WILL be independent of the initial angle theta. This is because of the law of conservation of energy. You can get this result by applying potential and kinetic energy rules.

The initial kinetic energy of the projectile is:

KEi = 0.5*m*u^2

At a point later on with height h, it will have potential energy PE = mgh. If you want to find the speed at that point, use:

KEF = KEi - PEf
0.5*m*vf^2 = 0.5*m*u^2 - mgh

This reduces to:
vf = SQRT(u^2-2gh)

You would have to use theta to determine the direction of this velocity vector, but not the magnitude.

I believe the problem you are running into is that if the projectile is fired vertically, it will actually achieve a velocity of 0 at the pinnaccle of flight. If you don't believe it, try it!

exactly. The y velocity will eventually become zero, at the projectiles highest point. Then the object will start to fall and the y velocity will become negative. You have to think in terms of y, not x