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Gravity in free fall equations

  1. Sep 30, 2015 #1
    Why is gravity negative in certain equations, such as
    y = yi + vi t + (1/2) ( − g ) t^ 2 , but positive in others like
    vf^2 =vi^2 +2gh ?

  2. jcsd
  3. Sep 30, 2015 #2
    It should be correct that the second equation you wrote is ##v_f^2=v_i^2+2(-g)h,## where my ##h## means the displacement of the object.
    If you write like that, you may regard ##h## as the height between ##x_i## and ##x_f##, where ##h## is always positive.
  4. Sep 30, 2015 #3

    Ray Vickson

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    It depends on whether increasing values of y are upward (i.e., measured up from the ground) or are downward (i.e., measured down from a point above the ground, such as from the roof of a tall building). If increasing y is upward, the acceleration of gravity is in the opposite (downward) direction, so appears with a '-' sign. If increasing y points downward, the acceleration of gravity points in the direction of increasing y, so would appear with a '+' sign.

    You can think about the second case yourself, but it essentially involves conservation of (total) energy.
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