Kinematics in Two Dimension-Niagra falls

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The discussion focuses on calculating the vertical distance at Niagara Falls where water's velocity vector points downward at a 75-degree angle. The initial horizontal speed of the water is 2.7 m/s, with zero initial vertical velocity and a constant acceleration of 9.8 m/s² due to gravity. To determine the vertical distance, the y-component of the velocity must be calculated to achieve the desired angle with the known x-component. The key is to find the appropriate y-component that results in a 75-degree angle for the resultant velocity vector. This calculation will ultimately lead to the vertical distance below the edge of the falls.
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Kinematics in Two Dimension-Niagra falls(not solved)

1. Suppose the water at the top of Niagra 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 75 degree angle below the horizontal.



3. Ok so i know the intial and final velocity in the x direction is equal to 2.7m/s with a 0 acceleration.. when it falls I assume that its initial velocity in the y direction is 0?(let me know if I am wrong here), and the acceleration to that is 9.8...now i need to figure out the time to get my Vfy, but I think i need My final position in order to calculate time. please help.
 
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Yes, you'd assume the y component of velocity is zero as it drops initially. What is key here is the resultant velocity vector being at 75 degrees below horizontal. You know the x component so you need to find the y component that will give that angle to the resultant. Then you can find the vertical distance.
 
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