Acceleration of Falling Object: -9.8 or +9.8?

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The discussion centers on the appropriate use of -9.8 m/s² or +9.8 m/s² for the acceleration of a falling object, emphasizing the importance of the chosen coordinate system. When defining upward as positive, gravity is represented as -9.8 m/s², while downward motions may be assigned positive values if downward is defined as positive. Consistency in assigning signs to vectors is crucial for solving problems accurately. The example of a ball thrown from a building illustrates how initial velocity and height can be incorporated into equations based on the defined coordinate system. Ultimately, the sign of the acceleration due to gravity depends on the direction designated as positive in the chosen reference frame.
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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?(-)
 
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Since gravity pulls objects downward that would be -9.8 m/s^2 independent of which way the ball is thrown.
 
It depends on your coordinate system. If distance is defined to increace up, g will be negative.
 
but if it says a ball is thrown *downward* at 25m/s that would be a positive g then??
 
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 +.
 
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.
 
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.
 
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.
 
thx u all !
 
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