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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...Asked by megan
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?
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Answered by
drwls
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.
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.
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