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Homework Help: Projectile Motion - Niagra Falls

  1. Dec 15, 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 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° angle below the horizontal?

    x direction

    a = 0 m/s

    v0x = 2.7 m/s

    vfx= 2.7 m/s

    x = ?

    y direction

    ay = -9.80 m/s2

    v0y = 0 m/s

    vfy = ?

    y = ?

    Other Info

    Angle = 75 degrees below horizontal

    t = ?

    2. Relevant equations

    Newton's laws of motion...but I'm not exactly sure which ones as it appears there are two variables missing from each one.

    3. The attempt at a solution

    I'm studying for my final this Friday. This question plagued me at the beginning of the term, I never got it. The class moved on, and so did I. But I'm looking at it again and am determined to get it, but am still drawing a blank.

    I'd like to find t as that would help significantly, but each equation that I've considered with it has another unknown variable as well.

    Kind of flustered...
     
  2. jcsd
  3. Dec 15, 2009 #2

    ideasrule

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    Homework Helper

    Imagine a water molecule falling off the edge of the waterfall. What's its vertical speed at time t? What's its horizontal speed? How are the vertical & horizontal speeds related to the 75 degrees?
     
  4. Dec 15, 2009 #3

    ideasrule

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    Homework Helper

    If you do it correctly, you WILL get t.
     
  5. Dec 15, 2009 #4
    tan-1 = (vy / vx) = 75 degrees
     
  6. Dec 15, 2009 #5
    Wait....

    vy / vx = tan 75

    vy = tan 75 (vx)

    vy = 10.0765 m/s



    vfy = voy + at

    vfy - voy / a = t

    t = 1.0282 s



    y = 1/2 (vfy + voy) t

    y = 1/2 (10.0765 m/s + 0 m/s) 1.0282 s

    y = 5.18 m ----> 5.2 m with sig figs


    Hrm....interesting.................
     
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