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 ?
-------
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?
--------
Quote:
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
|
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
-------
Quote:
|
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.
|
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!--
Quote:
|
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
|
Makes me wonder if there's a difference between moving vertically and horizontally o_o