That would be where the vertical velocity component Vy is related to the horizontal component Vx by
Vy/Vx = tan 55 = 1.428
Vy = 3.285 m/s (downward)
Calculate how far it must fall (Y) to reach that vertical velcoity component.
Vy = sqrt(2 g Y) = 3.285 m/s
2 g Y = 10.79 m^2/s^2
Y is less than 1 meter!
Suppose the water near the top of Niagara Falls has a horizontal speed of 2.3 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 55° angle below the horizontal?
I'm not sure what do with the angle.
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