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When I throw a ball upward on the earth, at a later time, that ball would be in the condition of $v=0$ (at the highest position) and then finally it would fall. If a ball falls, its direction of inertial force is upward. However, from it's initial condition to the $v=0$ condition (the highest position), or when its velocity is upward, the direction of movement and the direction of acceleration is opposite. In this condition, what is the direction of inertial force?

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    $\begingroup$ In what reference frame are you asking this? Note that there are no inertial forces in inertial reference frames. $\endgroup$
    – Sandejo
    Mar 7, 2021 at 2:02

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To an observer in the same freely-falling frame of reference of the object, the inertial force exactly cancels the gravitational force (principle of equivalence) .

And we know that the object falling is in a uniform gravitational field that acts downwards and at the rate given by $g$ and the value of this force is $mg$ for a mass $m$.

And as per that same Wikipedia article linked above:

"(Inertial mass) $\cdot $ (Acceleration) = (Intensity of the gravitational field) $\cdot$ (Gravitational mass)"

So the inertial force is in exactly the opposite direction of the gravitational force, and since the gravitational force acts toward the centre of earth, this means the inertial force will act away from the centre of the earth.

Note that an observer in this freely-falling frame of reference will detect no force at all.

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