3 ms·
Are you contesting the fact that a thrown ball's distance above the ground (y) at a particular time (t) relative to its initial velocity (v) is equal to y = vt
by albrewer 4y ago
Are you contesting the fact that a thrown ball's distance above the ground (y) at a particular time (t) relative to its initial velocity (v) is equal to y = vt + 0.5 * g * t ^ 2, (where g is the direction of acceleration due to gravity) which can very clearly a modeled by a parabola?
I get that the "error" is pretty low, but there are tons of equations in physics used every day that rely on the fact that the difference in some value is negligible (often things like the sine of an angle being approximately equal to the angle itself).
- deleted 4y ago[deleted]
- WastingMyTime89 4y agoNo but I get what they mean. If you view the Earth as a point system, you can use the general expression of gravity between two objects and you will indeed find an extremely elongated ellipse as a solution. Considering a parabola can be viewed as an ellipse with a focal point at infinity, I’m guessing the difference between the two solutions will be so small as to be negligible. They both depend on intentional inexactitudes (g is not a fixed vector, the earth is not a point system) which don’t matter at the scale. I wouldn’t personally view the elliptical solution as truer but I see how you could.
- albrewer 4y agoAccording to their link, the error at a human scale is about 1e-6 units on Earth. You need to be doing the math at velocities such that the object's apogee is over 100k feet / meters from the surface to be "off" by more than 1 unit.
- deleted 4y ago[deleted]