Suppose the water at the top of Niagara Falls has a horizontal speed of 2.59 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 67.9 ° angle below the horizontal?

Answer 1

Vx = 2.59
Vy = 9.8t

distance fallen = 4.9t^2

tan67.9° = 2.46 = Vy/Vx, so Vy = 2.46*2.59 = 6.37

so, the distance fallen is 4.9(6.37/9.8)^2 = 2.07 m