# -9.8 or just 9.8?

1. Sep 26, 2004

### saiyajin822

how can you tell when to use -9.8m/s^2 or +9.8 as your acceleration for a falling object? i have a problem that says " A ball is thrown upward from the top of a 25 m tall building." would that be -9.8 since the ball is being thrown up(+) and gravity is pulling it down?(-)

2. Sep 26, 2004

### Tide

Since gravity pulls objects downward that would be -9.8 m/s^2 independent of which way the ball is thrown.

3. Sep 26, 2004

### Integral

Staff Emeritus
It depends on your coordinate system. If distance is defined to increace up, g will be negative.

4. Sep 26, 2004

### saiyajin822

but if it says a ball is thrown *downward* at 25m/s that would be a positive g then??

5. Sep 26, 2004

### dekoi

Depending on whether you have assigned your point of reference and coordinate system as "Down" = "-" and "Up" = "+" or the other way around. If the ball is thrown downward, chances are all your values (or most) will be negative, so it is easier to make Down as a negative value on the coordinate system. Therefore, in that case, g will be +.

6. Sep 26, 2004

### Sirus

Right. As long as you are consistent throughout the problem in assigning signs to vectors according to whichever directions you defined as + or -, you will be fine.

7. Sep 26, 2004

### HallsofIvy

Staff Emeritus
However, most people, I think, would still choose their coordinate system with + upward and x= 0 on the ground. Then g= -9.8 m/s2 and, with the ball thrown downward, initial velocity -25 m/s.
That is, to determine the time in which a ball thrown downward at 25 m/s we would take initial speed -25m/s and initial height to be 25m and solve
-4.9t2+ 25t+ 25= 0.

Taking, as you suggest, the + axis downward we would probably take 0 to be at the top of the building, the ground to be 25 and solve 4.9t2+ 25t= 25.

8. Sep 26, 2004

### robphy

I prefer to do this: the vector $$\vec g$$ is the freefall acceleration vector pointing downward [toward the center of the earth]. Its magnitude is (using the convention of leaving off the arrowhead) $$g=9.8{\rm\ m/s^2}$$ (a non-negative quantity).

Now, the "sign" you seek is carried by the component of $$\vec g$$ along an axis of your choice.

If you call "upward" the "positive direction", then the component of acceleration along this axis is $a_y= - g$.

If you call "downward" the "positive direction", then the component of acceleration along this axis is $a_y= + g$.

9. Sep 26, 2004

thx u all !