Calculating Velocity Vector Below Niagara Falls Edge

[pookisantoki]

Suppose the water at the top of Niagara Falls has horizontal speed of 3.59m/s just before it cascades over the edge of the falls. At what vertical distance below the edge does the velocity vector of the water point downward at a 64.1 degree angel below the horizontal?

From this information I got
Ax=O
Vox=3.59
Y=?
Voy=?
Ay=-9.80
So i need to figure otu Voy can i do 3.59sin (64)??
And then what formula do I need to use??

 

[rl.bhat]

Here Vox remains constant. Voy = 0.
At a certain depth velocity is Vy.
Then Vy/Vox = tanθ. Find Vy and the depth.