3 ms·

The arc describes the full motion, and the solution we seek is when F(x)=0 which is when the ball hits the ground. There is no mathematical curve that starts at

The arc describes the full motion, and the solution we seek is when F(x)=0 which is when the ball hits the ground. There is no mathematical curve that starts at my hand at (x=0m, y=1m) and ends at the ground. We use the full quadratic curve just because its a suitable model for the motion, on the part of the motion we know the ball takes. The use of a quadratic to solve the throw is a mathematical model. We say that "the value x describing the when the ball lands must satisfy the quadratic equation F(x)=0" but that does NOT imply the opposite, which is "all x that satisfy F(x)=0 describe a valid motion of the ball." So when we get two answers, e.g. F(-1)=0 and F(15)=0 for the two points when the ball is at ground level, that means only this: if I had thrown the ball from ground level to follow the same curve land in the same place at x=15, then I would have stood 1m further back when I threw it. It does have physical meaning, but there is nothing curious about the physical meaning. This throw is symmetrical in time though, in the sense that if I throw the ball with the same speed in the opposite direction starting at x=15 then it will land exactly in my hand. (But the equation here is y=F(x) and not parametrized on time).

TeMPOraL 2y ago
I'd say 'jerf is correct here: https://news.ycombinator.com/item?id=43078998 https://news.ycombinator.com/item?id=43078998 The equation we use to describe the motion does not contain a term for the ground being there. That assumption exists outside of the model described by the equation, and you use that assumption after the fact to reject a solution that would otherwise be valid, and describe the movement of the ball that's pulled down by gravity. In this form, it doesn't really describe a proper orbit, just a trajectory of being pulled down by a constant force. I believe this would correspond to an infinitely-ish heavy object located infinitely far below. The proper equation that gives you an orbital curve has the force of gravity proportional to inverse-square distance and point at the center of the body, which is what makes it possible to describe a circular or elliptical motion this way. Parabolic orbits exist too, but they're interpreted as failed orbital capture - "object is moving so fast that it'll curve around and fly away to infinity before turning around and coming back". And in all cases, the solutions make physical sense (+/- infinity), on the assumption the trajectory doesn't cross the ground, as there's no term for it there :). If you want, you can describe the ground as another equation (or inequality), and solve the resulting system - it'll then be clear what exactly is it that rejects some of the solutions.

TeMPOraL 2y ago

I'd say 'jerf is correct here: https://news.ycombinator.com/item?id=43078998 https://news.ycombinator.com/item?id=43078998 The equation we use to describe the motion does not contain a term for the ground being there. That assumption exists outside of the model described by the equation, and you use that assumption after the fact to reject a solution that would otherwise be valid, and describe the movement of the ball that's pulled down by gravity. In this form, it doesn't really describe a proper orbit, just a trajectory of being pulled down by a constant force. I believe this would correspond to an infinitely-ish heavy object located infinitely far below. The proper equation that gives you an orbital curve has the force of gravity proportional to inverse-square distance and point at the center of the body, which is what makes it possible to describe a circular or elliptical motion this way. Parabolic orbits exist too, but they're interpreted as failed orbital capture - "object is moving so fast that it'll curve around and fly away to infinity before turning around and coming back". And in all cases, the solutions make physical sense (+/- infinity), on the assumption the trajectory doesn't cross the ground, as there's no term for it there :). If you want, you can describe the ground as another equation (or inequality), and solve the resulting system - it'll then be clear what exactly is it that rejects some of the solutions.