> in this regard everything is completely cogent and clear description of the forces (har har) in play.
Not everything you said, no. That's why I objected. If you want more detail, you said:
> The ball's trajectory will be a parabola. And at the highest point of that parabola the net vertical force on the ball is zero. It's where the force you exerted on it to send it up, and the force of gravity pulling it down eventually reach an equilibrium. Something inside of that ball would experience 0g at the moment when it was at its parabolic peak.
Actually, the net force on the ball (disregarding air resistance) is the same throughout the entire trajectory once it leaves your hand. (Here I'm taking the Newtonian view that considers gravity to be a force.) That's why the ball continuously accelerates downward by the same amount throughout the entire trajectory once it leaves your hand--which it has to in order for the trajectory to be a parabola. The force you exert on it to throw it upward stops as soon as it leaves your hand, so the only force thereafter is gravity. And since the force of gravity is not felt, the ball is in free fall, feeling 0 g, for the entire parabola.
So the part of your post that I quoted was not a "completely cogent and clear description of the forces"; it was a wrong description of the forces. That's why I corrected it.
Not everything you said, no. That's why I objected. If you want more detail, you said:
> The ball's trajectory will be a parabola. And at the highest point of that parabola the net vertical force on the ball is zero. It's where the force you exerted on it to send it up, and the force of gravity pulling it down eventually reach an equilibrium. Something inside of that ball would experience 0g at the moment when it was at its parabolic peak.
Actually, the net force on the ball (disregarding air resistance) is the same throughout the entire trajectory once it leaves your hand. (Here I'm taking the Newtonian view that considers gravity to be a force.) That's why the ball continuously accelerates downward by the same amount throughout the entire trajectory once it leaves your hand--which it has to in order for the trajectory to be a parabola. The force you exert on it to throw it upward stops as soon as it leaves your hand, so the only force thereafter is gravity. And since the force of gravity is not felt, the ball is in free fall, feeling 0 g, for the entire parabola.
So the part of your post that I quoted was not a "completely cogent and clear description of the forces"; it was a wrong description of the forces. That's why I corrected it.