Kinematics in Two Dimension-Niagra falls

[pstfleur]

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.

 

[Kurdt]

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.

 

[baba89]

I would approach this problem by breaking it down into smaller components and using the principles of kinematics to solve for the unknown variables. Firstly, we know that the initial velocity in the x-direction is 2.7 m/s and that there is no acceleration in this direction. Therefore, the velocity in the x-direction will remain constant throughout the fall.

Next, we can use the equation v^2 = u^2 + 2as to solve for the vertical distance below the edge where the velocity vector points downward at a 75 degree angle. In this case, our initial velocity (u) is 0 m/s, our final velocity (v) is unknown, our acceleration (a) is 9.8 m/s^2, and our displacement (s) is the vertical distance below the edge.

We can rearrange the equation to solve for v, which will give us the velocity at the point where the water is pointing downward at a 75 degree angle. Once we have this velocity, we can use trigonometry to find the vertical distance below the edge.

As for the time, we can use the equation v = u + at to solve for the time it takes for the water to reach this point. Again, we know the initial velocity (u) and acceleration (a), and now we have the final velocity (v) from our previous calculation.

Overall, the key to solving this problem is breaking it down into smaller components and using the equations of kinematics to solve for the unknown variables. With careful calculation and consideration of the given information, we can determine the vertical distance below the edge where the water points downward at a 75 degree angle.