Calculating Velocity Vector Below Niagara Falls Edge

[pookisantoki]

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??

 

[rl.bhat]

Here Vox remains constant. Voy = 0.
At a certain depth velocity is Vy.
Then Vy/Vox = tanθ. Find Vy and the depth.

 

[tomcue]

To calculate the vertical distance below the edge where the velocity vector points downward at a 64.1 degree angle, we can use the formula Vf^2 = Vi^2 + 2ad, where Vf is the final velocity, Vi is the initial velocity, a is acceleration, and d is the distance traveled. In this scenario, the final velocity is 0 m/s, the initial velocity (Vix) is 3.59 m/s, and the acceleration (a) is -9.80 m/s^2 (due to gravity). We also know that the angle between the initial and final velocity vectors is 64.1 degrees, so we can use trigonometric functions to determine the vertical and horizontal components of the initial velocity.

Using the given information, we can calculate the vertical component of the initial velocity (Viy) as Voy = Vix*sin(64.1) = 3.59*sin(64.1) = 3.12 m/s. Now, we can plug in these values into the formula Vf^2 = Vi^2 + 2ad and solve for d:

0^2 = (3.59)^2 + 2*(-9.80)*d

0 = 12.88 - 19.6d

19.6d = 12.88

d = 12.88/19.6 = 0.657 m

Therefore, the water will travel 0.657 meters vertically below the edge of Niagara Falls before the velocity vector points downward at a 64.1 degree angle.