> That is just not going to happen in (classical) reality, though. Because once you properly set the initial state of the ball (force=velocity=0, or any other values), then the solution becomes unique and that’s it, there is one possible trajectory.
Not true, and the trick is very simple, you just contrive a force field that admits x''(0) = 0 but some non-zero higher derivative. Indeed, if you look at the proposed solution r = 1/144 (t - T)^4 for t > T, you'll see that:
r(T) = 0
r'(T) = 0
r''(T) = 0
r'''(T) = 0
r''''(T) is ill-defined (but Newtonian mechanics AFAIK says nothing to forbid this)
Not true, and the trick is very simple, you just contrive a force field that admits x''(0) = 0 but some non-zero higher derivative. Indeed, if you look at the proposed solution r = 1/144 (t - T)^4 for t > T, you'll see that:
r(T) = 0
r'(T) = 0
r''(T) = 0
r'''(T) = 0
r''''(T) is ill-defined (but Newtonian mechanics AFAIK says nothing to forbid this)