Suppose the water near the top of Niagara Falls has a horizontal speed of 2.7 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 72° angle below the horizontal?

Answer 1

The water fows at that angle when the ratio Vy/Vx = tan 72 = 3.08

Vx remains 2.7 m/s

Vy = sqrt(2Y/g) = 3.08*2.7 = 8.3 m/s

Calculate Y (the distance the water has fallen) using that value of Vy.