Help Kinematics in 2d Question

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The discussion centers on solving a kinematics problem involving the motion of water at Niagara Falls. The horizontal speed of the water is 1.23 m/s, and the goal is to determine the vertical distance where the velocity vector points downward at a 49.4-degree angle. The relationship between the vertical and horizontal components of velocity is established using the tangent function, leading to the equation tan(49.4) = Vy / 1.23. Subsequently, the vertical displacement is calculated using the kinematic equation (Vy)^2 = u^2 + 2gy, where the initial vertical velocity (u) is zero.

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  • Knowledge of the kinematic equation for vertical motion
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Help! Kinematics in 2d Question!

Suppose the water at the top of Niagara Falls has a horizontal speed of 1.23 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 49.4 degrees angle below the horizontal?
 
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The angle is given as 49.4degrees...
And you know that tan of an angle will be y/x, and you're given the x velocity component, 1.23m/s...
Now, you can form an equation...
tan 49.4 = Vy/1.23
Solve it for Vy... (Vy is that final vertical velocity)
Then, to find the vertical displacement, use the formula (Vy)^2 = u^2 + 2gy given Vy, g and u. (It's a waterfall... so vertical velocity is assumably initially 0m/s)
Solve it for y. :}
 
Oh..

Hey thanks for your help man ... BUT ... i got it wrong ..andi failed ..
 

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