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B Is the value of g negative or positive?

  1. Dec 12, 2016 #1
    I know that in formus there are many answers about this question .I just want the value of g (will it be positive or negative) in this example which I am going to give , I threw a ball vertically upward the value of g at its max height (vi=0) ,what will be the value of g +10 or -10 m/s^2. I think it can be both if you consider the cartisean plane's cordinate , it depends upon refrence.
     
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  3. Dec 12, 2016 #2

    Svein

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    g is a vector (pointing to the center of the Earth).
     
  4. Dec 12, 2016 #3

    CWatters

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    What sign did you use for the initial velocity? ditto the displacement/height?
     
  5. Dec 12, 2016 #4
    This means with the sign of Vi you decide the value of g to be positive or negative.
     
  6. Dec 13, 2016 #5

    CWatters

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    Correct. It depends which direction (up or down) you define as positive.
     
  7. Dec 13, 2016 #6

    jtbell

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    In all the textbooks that I remember using (there may be exceptions elsewhere!), g is defined as a positive number, the magnitude of the (vector) acceleration of a freely-falling object near the earth's surface. It is used with a + or - sign depending on the coordinate system.

    If position, velocity, and acceleration are defined to be positive in the upwards direction and negative in the downwards direction, then the (y-component of) acceleration of a freely-falling object is ##a = -g## (more precisely ##a_y = -g##) and its position as a function of time is ##y = y_0 + v_{0y} t + \frac 1 2 a_y t^2 = y_0 + v_{0y} t - \frac 1 2 gt^2##.

    If position, velocity, and acceleration are defined to be positive in the downwards direction and negative in the upwards direction, then the (y-component of) acceleration of a freely-falling object is ##a = +g## (more precisely ##a_y = +g##) and its position as a function of time is ##y = y_0 + v_{0y} t + \frac 1 2 a_y t^2 = y_0 + v_{0y} t + \frac 1 2 gt^2##.
     
  8. Dec 13, 2016 #7

    sophiecentaur

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    You start with a diagram. You put xy axes on it. That defines the sign of your y (vertical) position. If you choose to call 'up' positive, then g will be negative. If you choose 'down' to be positive (say you are dropping things off a bridge) then your g will be positive. It doesn't matter what you choose as long as you stick with the signs throughout . :smile:
    Soon your diagram can be in your head.
     
  9. Dec 14, 2016 #8
    I got it thanks .
     
  10. Dec 19, 2016 #9

    David Lewis

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    The direction of the ball's motion does not determine whether g is positive or negative.
    If you call the directional sense of gravitational force negative, then g will also be negative.
     
  11. Dec 20, 2016 #10
    I always tell my students to double check that the real effect of their equations is a downward acceleration.

    Best practice is to define your coordinate system on the paper with y clearly pointing up. But if you point y down, gravitational acceleration is going to be positive.
     
  12. Dec 20, 2016 #11

    A.T.

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    It basically boils down to what you mean by "g".
     
    Last edited: Dec 20, 2016
  13. Dec 20, 2016 #12

    ZapperZ

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    My students often hated me for forcing them to do this, but in at least 3 or 4 instances, I make them solve a problem by placing the origin in another location, and changing the direction that they call positive. I do this for projectile motion and simple Coulomb's law problems.

    The whole exercise was meant to show that the location on where we define to be "zero" and the orientation of our coordinate axis are really arbitrary, and that no matter how we do this, it should not affect the final answers. Mother Nature doesn't give a hoot on which direction we call "positive" and "negative", and thus, your answer shouldn't either.

    Zz.
     
  14. Dec 20, 2016 #13

    CWatters

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    ...as long as they are consistent in their definition.
     
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