# Is the value of g negative or positive?

• B
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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Svein
g is a vector (pointing to the center of the Earth).

CWatters
Homework Helper
Gold Member
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.
What sign did you use for the initial velocity? ditto the displacement/height?

Ali Hamaiz
This means with the sign of Vi you decide the value of g to be positive or negative.

CWatters
Homework Helper
Gold Member
Correct. It depends which direction (up or down) you define as positive.

jtbell
Mentor
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##.

Ali Hamaiz
sophiecentaur
Gold Member
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 .

Ali Hamaiz
I got it thanks .

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.

Dr. Courtney
Gold Member
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.

A.T.
g is a vector (pointing to the center of the Earth).
g is defined as a positive number, the magnitude of the (vector)
It basically boils down to what you mean by "g".

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Redbelly98
ZapperZ
Staff Emeritus
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.

CWatters
CWatters