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Freefall acceleration

  1. Aug 30, 2011 #1
    I worked two similar freefall problems with the same coordinate system, but had different signs for acceleration on each solution. I thought that if we say +y is up, then a = -9.8 m/s², so I'm confused as to why a = +9.8 m/s² for the second problem.

    Problem 1: Rock thrown from 30 m cliff with initial velocity of 12 m/s. Find v when it hits ground.
    Coordinate system: +y is up and 0 is on ground. (so a = -9.8 m/s²)
    Solution: vf² = v0² + 2a(x - x0) = (12 m/s)² + 2(-9.8 m/s²)(0 m - 30 m) = 732
    vf = 27.1 m/s
    This makes sense to me because if a = +9.8 m/s², I'd have to take the square root of a negative.

    Problem 2: A rock passes a 3m window in 0.4 s. At what height above the window was it dropped from.
    Coordinate system: +y is up, zero at bottom of window. (so a = -9.8 m/s²)
    Solution: Let d be the distance from the drop point to the top of the window. First find velocity at top of window. Assume rock dropped from rest, then find d.
    (x - x0) = v0t + ½at²
    (0 m - 3 m) = 0.4v0 + ½(-9.8 m/s²)(0.4 s)²
    v0 = -5.54 m/s
    vf² = v0² + 2a(x - x0)
    (-5.54 m/s)² = 0 + 2(-9.8 m/s²)[(d + 3) m - 3 m]
    d = -1.57 m
    Which doesn't make sense physically, because it would be below the bottom of the window.So I tried it with a = +9.8 m/s², I get -9.46 m/s for velocity at the top of the window and d = 4.56 m.
     
  2. jcsd
  3. Aug 30, 2011 #2

    tiny-tim

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    welcome to pf!

    hi willcrys84! welcome to pf! :smile:
    it doesn't make any difference whether it's 12 m/s up or 12 m/s down, does it? :wink:
    x0 is the top, so (x - x0) is negative :wink:
     
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