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I mean if, for example, I throw a rock upward, its acceleration will always be $-g$ or it will be $-g+a$ because I apply a force on the object when I throw it? (without considering friction)

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  • $\begingroup$ It's $-g+a$ only a few fractions of a second, really. $\endgroup$
    – DanielC
    Mar 22, 2021 at 10:20

2 Answers 2

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Its acceleration will be $-g+a$ while you are applying the force on it, but it will be only $-g$ when you end applying the force. If there were friction, the object would experience an additional acceleration on both cases.

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  • $\begingroup$ It is not clear to me what you mean by $a$ here. $\endgroup$
    – B. Brekke
    Mar 22, 2021 at 12:24
  • $\begingroup$ The acceleration produced by the launching force $F$, i.e. $a=\frac{F}{m}$, with $m$ the mass of the object. $\endgroup$
    – AFG
    Mar 22, 2021 at 12:40
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Remember that in classical mechanics gravity is a force and not an acceleration.

The acceleration of your rock will be the second derivative of its position in time.

The force acting on your rock will be the sum of gravity and all the other forces that may act on it.

Newton's lav $F=ma$ tells you how to connect force and acceleration, once you know one of the two. You have to remember that $F=ma$ is a principle, not a definition.

You confusion arises from the fact that, due to its particular form, it happens that the acceleration that gravity impresses on objects will be always $g$ (close to the Earth surface) regardless of their mass $m$, if no other forces are present.

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