Water at Niagara Falls cascading at 49.4 degrees below horizontal

[Morokana]

Help! Kinematics in 2d Question!

Suppose the water at the top of Niagara Falls has a horizontal speed of 1.23 m/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 49.4 degrees angle below the horizontal?

 

[Pseudo Statistic]

The angle is given as 49.4degrees...
And you know that tan of an angle will be y/x, and you're given the x velocity component, 1.23m/s...
Now, you can form an equation...
tan 49.4 = Vy/1.23
Solve it for Vy... (Vy is that final vertical velocity)
Then, to find the vertical displacement, use the formula (Vy)^2 = u^2 + 2gy given Vy, g and u. (It's a waterfall... so vertical velocity is assumably initially 0m/s)
Solve it for y. :}

 

[Morokana]

Oh..

Hey thanks for your help man ... BUT ... i got it wrong ..andi failed ..