[Coda]

Originally Posted by Potironette

An answer! Thanks!

For the accelerating downward, I'm pretty sure that when the platform accelerates down, if it is accelerating at gravity, then then a ball is dropped, the ball doesn't fall for the person inside, though it's falling to the person outside at normal speed. If the platform accelerates downward faster than gravity, then the person inside sees the ball floating upwards while the person outside sees it falling normally. If the platform accelerates downwards slower than gravity, then the person inside sees the ball fall slower down, I think, but either way the person outside sees the ball falling normally.

Correct: If the platform is accelerating downward at the same rate as gravity, then the ball will always remain at the same distance from the platform.

It should be noted, of course, that a person standing on the platform would levitate off of the platform if it's descending faster than gravity! That's what seat belts are for. :P

Actually, maybe I shouldn't even think about this since once the ball is dropped during upward acceleration, the ball goes up according to velocity and so it would be extremely complicated to find when it would hit the ground?

It's not actually as complicated as you might think. You just have two equations -- one describing the ball's motion according to gravity, and one describing the platform's motion under its own power -- and you solve them as a system to determine when those paths intersect.

For example:

Ball: y = 10 - g*t^2 (that is, starts at 10m in the air and falling with gravity)
Platform: y = 0 + 15*t^2 (that is, starts at 0m on the ground and goes up at 15m/s)

Since you're asserting that you want to know when y is the same for both of them, you can set them equal to each other and solve for t:

10 - g*t^2 = 0 + 15*t^2
10 = 15*t^2 + g*t^2
10 = (15 + g)*t^2
10 / (15 + g) = t^2
0.068 ~= t^2
0.261 ~= t

So the ball hits the platform after approximately 0.261 seconds. You can then plug the time back into the ball's position equation:

y = 10 - g*0.261^2
y = 10 - 0.668
y = 9.332

So when the ball hits the platform, they're 9.332 meters above the ground. (Which means that the platform is shooting up REALLY fast and I should have picked a smaller number. XD )

Meanwhile, if you want to find the position of the ball relative to the platform you just have to subtract:

y' = Ball(t) - Platform(t)
y' = 10 - g*t^2 - 15*t^2
y' = 10 - (15 + g)*t^2

(Of course, this is only valid until they collide.)

Edit: After thinking about it for a while, that answer is really helpful! I forget that I can see the elevator moving up and the ball moving up as it is also falling down from the outside XD

Glad I could help!

EDIT: The above techniques still work if you throw the ball upward instead of just dropping it. The math takes a few more steps because it's in the form y = y0 + vt + at^2 but it's totally manageable.