Calculating Velocity Vector Below Niagara Falls Edge

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SUMMARY

The discussion focuses on calculating the vertical distance below the edge of Niagara Falls where the water's velocity vector points downward at a 64.1-degree angle. The horizontal speed of the water at the top is 3.59 m/s, and the vertical acceleration due to gravity is -9.80 m/s². The participants derive that the vertical component of velocity (Voy) can be calculated using the formula Voy = Vox * sin(θ), where Vox is the horizontal velocity. The relationship between vertical and horizontal velocities is established using Vy/Vox = tan(θ) to find the vertical distance.

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pookisantoki
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Suppose the water at the top of Niagara Falls has horizontal speed of 3.59m/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 64.1 degree angel below the horizontal?

From this information I got
Ax=O
Vox=3.59
Y=?
Voy=?
Ay=-9.80
So i need to figure otu Voy can i do 3.59sin (64)??
And then what formula do I need to use??
 
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Here Vox remains constant. Voy = 0.
At a certain depth velocity is Vy.
Then Vy/Vox = tanθ. Find Vy and the depth.
 

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