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Homework Help: -9.8 or just 9.8?

  1. Sep 26, 2004 #1
    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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  3. Sep 26, 2004 #2


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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.
  4. Sep 26, 2004 #3


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    It depends on your coordinate system. If distance is defined to increace up, g will be negative.
  5. Sep 26, 2004 #4
    but if it says a ball is thrown *downward* at 25m/s that would be a positive g then??
  6. Sep 26, 2004 #5
    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 +.
  7. Sep 26, 2004 #6
    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.
  8. Sep 26, 2004 #7


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    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.
  9. Sep 26, 2004 #8


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

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

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

    If you call "downward" the "positive direction", then the component of acceleration along this axis is [itex]a_y= + g[/itex].
  10. Sep 26, 2004 #9
    thx u all !
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