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If the outsider is at rest relative to the moving platform, and the moving platform is in fact accelerating upward, then yes, in the end, the ball does accelerate upward in the end. The observer will see the ball accelerate downward until it collides with the platform, then accelerate upward as the platform pushes it back up.

You can determine the position of the ball at any point in time if you know where it started, where the platform started, and the acceleration of the platform.

The more interesting question is if the platform is accelerating DOWNWARD independent of gravity, because then the ground is falling away from the ball at a different rate than the ball is itself falling.

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.

My problem with the accelerating upward is that I can't visualize it from an outsider's point of view. For the accelerating downward platform, it'll be going down no matter what. However, for the upward accelerating platform, the path the ball takes from the outside is confusing '~'.

If the platform accelerates up at the value of gravity(?), then the ball goes up, then hits the ground where it started? I have no clue.
If the acceleration of the platform is up less than gravity, then the ball goes up, then hits the ground lower than where it started?
If the acceleration of the platform is up greater than gravity, then the ball goes up, then hits to ground higher than where it started?

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?

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

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

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.)

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.

Ohh, so on a moving platform since the ball is being dropped the velocity starts at zero! I didn't realize that, woops X'D. Also, thanks a lot especially for doing the math part! I have a list of equations from school, but had no idea which one relavant, and what the different variables would stand for. Other than that, I wasn't particularly motivated to try to go through the equations :x.

Although, why isn't the
ball y = 10 m - (1/2)*g*t^2 and the
platform y = 0 + (1/2)*(15 m/s^2)*t^2 ?

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So..
If I put an extremely long ruler vertically up to the sky, y would be a point on the ruler, and that is why when the y of the ball and the y of the platform are equal, they hit each other?
If I'm getting the bit about y being from the view of the outsider wrong, then everything that follows will be wrong :x

Ball: y = d - g*t^2
Platform: y = d, + a*t^2

(where d and d, are also points of the metaphorical ruler...or they are both according to the person on the platform's perspective.)

d - g*t^2 = d, + a*t^2
d - d, = a*t^2 + g*t^2
d - d, = ag(t^2)
(d - d,)/(ag) = t^2
sqrt((d - d,)/(ag)) = t

Ball: y = d - g*sqrt((d - d,)/(ag))
or
Platform: y = d, - a*sqrt((d - d,)/(ag))
((where the y is how high up the building the platform or ball is when they meet))

And, based on this... in order for the ball dropped to hit the platform at the location it fell from...

Ball: y = d - g*t^2
Platform: y = d, + a*t^2
y = d ? Or in
which means
Ball: d = d - g*t^2
0 = -g*t^2
meaning it's not possible for that to happen :o?

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And for the position of the ball you mean it's possible to find out the distance between the ball and the rising platform given a random time?
Ball: y = d - g*t^2
Platform: y = d, + a*t^2
(where d and d, are either on the metaphorical ruler or according to the person on the platform)
(where the y of the ball is not the same as the y of the platform. Where the y of the ball is where the ball is and the y of the platform is where the platform is.)
y' = How high the ball is - how high the platform is
y' = (d - g*t^2) - (d, + a*t^2)
y' = d - g*t^2 - d, - a*t^2
y' = d - d, + - (a + g)*t^2

-------
Ball: y = d + vt + gt^2
Platform: y = d, + at^2
d + vt + gt^2 = d, + at^2
d - d, = at^2 - gt^2 - vt
--I have no clue how to solve for t, but I can see it's possible!--

Makes me wonder if there's a difference between moving vertically and horizontally o_o

this is so over my head

It couldn't fall very well if it was at head-height, of course. :D

Beeeeeeecause I remembered wrong. ^^() I was working from memory and I don't have to worry about that particular division when I'm working on game code because I'm fiddling with the constants ANYWAY until I find a result that feels playable. You're correct, and you should modify my examples appropriately.

Yes, exactly. y is measuring the positions of the objects from the perspective of an independent observer at rest.

No, solving that last step for t results in t = 0, which means that given the initial conditions you put into the system, they start off in contact with each other.

Which is, in fact, what you did by using d in both of those position equations -- what you're calling d is the initial position of the object, and you've used the same position for both of them. You'll notice in my example that I had the ball starting at 10 and the platform starting at 0.

Yes, that's what I mean. You can determine that, as long as that time is between 0 and the point of contact. After the point of contact, those equations would be describing what would happen if they hadn't actually hit each other (for example if they passed through each other or if the ball was dropped BESIDE the platform instead of OVER it).

First, let's fix the problem with the two objects starting in the same place by calling the ball d and the platform e, and putting the missing 1/2 in.

To solve it, first you'd simplify:
d - e = (1/2)(a - g)t^2 - vt

Then you want to get 0 on the left:
0 = (1/2)(a - g)t^2 - vt - d + e

Then since that's in the form 0 = ax^2 + bx + c, you'd use the quadratic equation:
x = (-b +/- sqrt(b^2 - 4ac)) / (2a)

Substituting in the stuff from the equation, then that's:
t = (v +/- sqrt(v^2 - 4*((a - g)/2)*(-d + e))) / (2*(a - g)/2)

It looks scary, but a, g, v, d, and e are all constants, so you can plug it into a calculator.

This will return two values for x (one when you use +, one when you use -) that are solutions to the equation.

Nope, there's not. The only thing that makes motion "vertical" is that it's parallel to the direction of gravity. All you have to do is replace g with the acceleration due to gravity in the direction of motion; for horizontal motion, that means g is 0.

I split this off into its own thread because this isn't the first time I've helped people with stuff like this and it makes sense to give it some more visibility if I'm going to do this more often. :P

A new thread! Makes me wonder if I should change the first post XD. Thanks for helping me with this! School's definitely not asking me to answer these questions, but I get pretty confused when I have questions based on what school is telling me.

So.. something like this can't happen? :

I used " d, " with the comma for the platform, since I was worried I'd forget that I was trying to think about positions.

-------------------------

Ball: y = d - (1/2)g*t^2
Platform: y = e + (1/2)a*t^2
y = d so the ball ends up where it started
Ball: d = d - g*t^2
0 = -g*t^2
meaning that for the ball to hit the ground where it began, no time has passed, since knowing only the initial and final position, it's that the ball has not moved at all.
But..
Platform:
d = e + r where r is the distance between d and e
d = e + (1/2)a*t^2
e + r = e + (1/2)a*t^2
r = (1/2)a*t^2
In this case the time is not zero because this equation has more information on what the ball is doing?

That means that if somehow, there was an accelerating box on the ground moving all around earth somehow, and there was a person in it, the person would feel that gravity was whatever direction opposite where the person was accelerating in? Would they think the ground was the side of the box accelerating towards them and they'd feel like the box was moving to side pushing them against the side of the box on the ground :o? Actually, this reminds me sort of the NASA centrifuge I keep hearing mentions of in class, though I'm not sure how people feel inside there.

That could happen, but that's not what the equations you set up represent. That would be more like:
Ball: y = 10 - v*t - (1/2)g*t^2
Platform: y = 0 + (1/2)a*t^2

Oh, I thought the comma was a typo because you didn't use it anywhere else. ^^() Usually when you want to distinguish two related variables, you use a prime (') or a subscript (but Trisphee doesn't have bbcode for that; sometimes it's typed with _, for example using "v_0" to mean "initial velocity").

The main problem here is that you're lacking an initial velocity term. Without an initial velocity, the ball will never go up, which means the only time it can be at its starting position is at the start.

If you DID have a velocity term in there, then the equation would have two solutions (as I described with the quadratic formula), representing the two different times the ball was at position d.

This isn't quite representing the same system -- the previous system is a special case of this one where r = 0, but this system of equations can represent more possibilities.

Not quite. The person would feel gravity pointing towards the surface of the Earth, and they would ALSO feel ANOTHER force coming from the side of the box pushing them around, in the same direction as the box's acceleration. (If the box were maintaining a constant speed, they wouldn't feel that.) You can add the force vectors together (this is a two-dimensional operation) to determine the net force that the person would feel. You could theoretically put a diagonal platform inside that box perpendicular to that net force vector, and the person could stand on it.

If you were standing on the floor of a rocket steadily accelerating through outer space, then the back wall of the cockpit would feel like the "floor" from the perspective of the occupant because of the direction of the acceleration.

The centrifuge is a substantially more complicated bit of math because it's a rotating system, and one of the forces comes from the fact that the end of the centrifuge ISN'T moving relative to the passenger so it's pushing back on the passenger to keep him from flying out the end from sliding down the side. At that point you're getting into calculus if you want to derive the forces from first principles; in practice, that work has already been done for you and you work with a different set of force equations for rotating systems.

Ohh, so it can happen because if y = 10, then
0 = -v*t - (1/2)g*t^2
v*t = - (1/2)g*t^2
v = - (1/2)gt
-v = (1/2)gt
and the velocity will be a number so time can pass! And the negative is probably because gravity and the velocity upward are not going in the same direction?

So if y = d
Ball: 0 = -v*t - (1/2)g*t^2
Platform: d = d' + (1/2)a*t^2
at' = v where t' is how much time the platform has been accelerating in total
And then to plug that into 0 = -v*t - (1/2)g*t^2
at't = - (1/2)gt^2
a = (-(1/2)gt^2)/(t't)
So... given the time the platform has been moving and given the time it took for the ball to hit the platform, it's possible to find an acceleration for which the ball falls lands on the platform where it started? It looks a lot less convenient than I'd imagined in the beginning!
Uh, granted I have no clue if this is messed up or not, or if it's a quadratic equation. It doesn't really look like one, I think.

At this point I think it's starting to go in over my head X'D. I don't really understand how equations relate to each other, or what systems are.

Ohh that's really cool! How much of a stretch would it be to say that in the moment a person is being pushed or pulled, their idea of gravity is changed and so they lose their balance?

As for the centrifuge, I sadly haven't learned a thing about rotating systems yet, or calculus '~'.

Coda, you just became 10000x sexier to me

Um... actually the negative is because I wasn't paying attention to the signs I was using in other places, and it should have been a +. But you derived the correct conclusion from it anyway!

t is always how long the system has been accelerating in total. You don't need to define a t'. You might choose to define a t_1 or something that specifies a particular time in the sequence (perhaps you might designate it to be the time when the ball meets the platform) but if you do that you would plug it in as a value for the t variable instead of assigning it to the other side of the equation.

I don't know how old you are or what class you're taking, so I don't know what your background is or what you've already studied. :P

The short version is that if you have multiple equations, then the only relationship between them is that they all have to stay true all the time, and any variable that appears has to have the same value in all of the equations. (Even if you don't know what that value IS.)

If you can solve any of the equations for one of the variables in it, then you can substitute that solution in place of that variable in any of the other equations. Usually when you're solving a system of equations your goal is to do this iteratively until you produce one that's entirely expressed using a single variable.

THAT starts getting into terms with a t^3 factor, which is called "jerk" and describes the rate of change of force. Humans are pretty good at dealing with forces that change slowly, and you can keep your balance reasonably well if you're given time to adjust your stance.

But if you're saying the box suddenly starts accelerating, then yeah, the person is PROBABLY going to fall over.

I didn't expect you to have, which is why I didn't go into any further detail. :P I mentioned that it COULD be done and left it at that.

Sorry, ladies, I'm married.

Man, things are so tough I can't even admire a sexy brain in another continent :(

Thanks a lot for all the answers! I don't really remember the math too well, but at least I get that it's possible for the dropped ball to return to the platform accelerating upwards at the point where the ball was dropped, and that for that to happen, it requires knowledge of the velocity of the ball when it was dropped, and that the velocity depended on the acceleration. Although, I'm confused about the t variable. Since it's how long the system has been accelerating in total, what is how long it takes for the platform to hit the ball :o?