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When we throw an object up into the air, ignoring air resistance, etc, we define acceleration to be -9.8 m/s^2. When it goes down after its journey up, like a parabola, do we define the acceleration as 9.8 m/S^2, OR still the same as -9.8m/s^2?

I know that acceleration due to gravity is always constant, but would it be wrong to say it like the question i have above?

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  • $\begingroup$ How do you calculate acceleration in the first place? What's the definition? Calculate what is happening to the object at different times and see what you get. $\endgroup$
    – BowlOfRed
    Sep 7 '20 at 22:46
  • $\begingroup$ We don’t define acceleration to be a particular value. We define it to be the rate of change of velocity with time. $\endgroup$
    – G. Smith
    Sep 7 '20 at 22:59
  • $\begingroup$ The acceleration due to Earth’s gravity is not constant. Its magnitude changes with altitude, and its direction changes with latitude and longitude. It is approximately constant over a small region. $\endgroup$
    – G. Smith
    Sep 7 '20 at 23:02
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The simple answer is yes it does stay the same $-9.8ms^{-2}$ because the acceleration is always pointing downwards towards the earth so we keep the minus. if you change coordinate systems you will need to change the result but with being close to the earth allowing to approximate the acceleration due to gravity to a constant which is in this case $9.8ms^{-2}$ with the downward direction assumed to be the negative direction we will keep $-9.8ms^{-2}$ for all constant acceleration equations in the earth's gravitational field.

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The sign used for expressing the acceleration due to gravity totally depends on you. If the acceleration is in the opposite direction of your velocity then you can take any of them as positive or negative . By convention , since the acceleration is in downward direction and it opposes the motion we take it as negative.

When the ball comes down , its velocity and acceleration due to gravity both are in the same direction i.e the velocity is increasing with time . So here you can take both of them to be positive or both of them as negative (as per your convenience).

It doesn't matter but it's better to use these quantities with assigned sign convention .

Hope it helps 🙂.

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The sign conventions then would put displacement as negative when your object is coming down. so when you use equations like $s=ut+\frac{1}{2}at^2$ what you would substitute in is -9.8$m/s^2$ and -s where -s is the displacement downwards

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  • $\begingroup$ sign conventions being that you take down to be positive and up to be negative $\endgroup$
    – F.N.
    Sep 8 '20 at 4:14
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Acceleration is the rate of change of velocity with respect to time, and velocity is the rate of change of height with respect to time (if we are just considering vertical motion).

If you define height as increasing as you go up (which is the usual convention) then in ballistic motion height increases to a maximum value and then decreases. So velocity starts out positive, and decreases both on the way up and on the way down - remember that a velocity of $-10$ m/s is less than a velocity 0f $-5$ m/s. So the acceleration due to gravity is negative throughout the motion.

On the other hand, if you define height as decreasing as you go up then in ballistic motion height decreases to a minimum value and then increases. So velocity starts out negative, and increases both on the way up and on the way down. So acceleration is positive.

In other words, once you choose a direction for increasing height, the sign of the acceleration due to gravity is then fixed and stays the same throughout any ballistic motion.

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