In free-fall acceleration, how do I figure when g is positive or negative?

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Elis
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In free-fall acceleration, how do I figure when "g" is positive or negative?

Homework Statement


My textbook is confusing me a bit. In general, when would "g" be positive and when would be negative? I thought it was when the particle was falling downward, it was positive and when it's going upward it's negative, but I seem to have that incorrect.

Somebody also told me that if I have y-coordinate with the bottom portion as positive and the top portion as negative, g would then be positive. However, she said it would also be positive if the reverse occurred. I don't quite understand that.

Could someone please explain this?


Homework Equations


No, eq. General question.


The Attempt at a Solution

 
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It depends on what system of coordinates you choose to adopt. If the positive direction of your y-axis is up, then g will be negative. If your y-axis is pointing down, then g is positive.

All you need to make sure is that your sign assignments are consistent, so if your y-axis is pointing up, then a throw up against gravity would be described by: [tex]y(t)=vt-\tfrac{1}{2}gt^2[/tex]

If, however, I were to choose a y-axis pointing down, then the motion of the thrown object would be described by:
[tex]y(t)=-vt+\tfrac{1}{2}gt^2[/tex]
 
Last edited:


Elis said:

Homework Statement


My textbook is confusing me a bit. In general, when would "g" be positive and when would be negative? I thought it was when the particle was falling downward, it was positive and when it's going upward it's negative, but I seem to have that incorrect.

It is a matter of convention whether you defined up as positive, or negative.

The acceleration is always downwards.

When an object is thrown upwards, its velocity is continually changing. You can think of this as adding small increments of velocity. If something is moving up at 20 m/s, and one second later it is moving up at 10.2 m/s, what is the velocity that has been added? Have you added a velocity going up, or added a velocity going down?

Cheers -- sylas
 


I understand that now; thank you very much for your help.