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Finding a vertical distance of niagara falls?

  1. Jun 8, 2009 #1
    1. The problem statement, all variables and given/known data
    Suppose the water at the top of Niagara Falls has a horizontal speed of 1.42 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 61.8 ° angle below the horizontal?


    2. Relevant equations

    Not sure

    3. The attempt at a solution
    Initial velocity(u) of Niagara falls = 1.42 m/s
    Angle(θ) at which it strikes the ground = 61.8o
    Formula used: tan(61.8) = sq.rt(2*9.8*x/1.42)
    I solved for x and got .252 m but the answer is wrong. Now I have no idea on what to do. Any help would be appreciated. Thank you.
     
  2. jcsd
  3. Jun 8, 2009 #2
    Remember the following: the horizontal component of the velocity will never change (assuming we are only dealing with gravity). So what you need to do is find the vertical velocity that will make the velocity vector point at the desired angle. Then recall from kinematics:

    vf2=vi2+2*g*d

    vf is the final velocity
    vi is the initial velocity
    g is the acceleration due to gravity
    d is the distance travelled

    of course this is all for the vertical components, and vi is zero, so you just have to solve for d.
     
  4. Jun 8, 2009 #3
    okay but how do i find the vertical distance, I am barely on my 4th day of physics so I really do not know much. But I got these hints off my book, but I have no idea how to translate what it says into a drawing.

    Sketch an arrow that represents the velocity vector after the water has fallen some vertical distance. It should be angled downward. Construct a right triangle whose hypotenuse is the velocity vector of the water, and whose sides are the vertical and horizontal components of the velocity, and . The horizontal component of the velocity is known. Use the Pythagorean theorem to solve for the magnitude of the vertical component of the velocity.

    Once the vertical component of the velocity is known, use the equations of kinematics (table 3.1) to solve for the vertical displacement from the top of the waterfall. The magnitude of the vertical displacement is equal to the distance below the edge of the waterfall.
     
  5. Jun 9, 2009 #4
    That hint your book gives you is essentially what I have told you, except I go farther and give you the equation. So you are having trouble drawing the diagram. Ok, well consider that you have two components to the velocity of the water falling from the cliff, a component in the horizontal direction (IE parallel to the ground) and a component in the vertical direction (IE perpendicular to the ground.) So draw two lines connecting at a point, but perpendicular to eachother. These are the components of the velocity of the water. Now join them with a third line to make a triangle, and this is the velocity vector. So now all you need to do is set the appropriate angle to 61.8 degrees, and this will set the length of the vertical component, because the horizontal component is already known, and you have an angle at 90 degrees as well.
     
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