Yes, the ball would land in exactly the same spot, whether robot or person. The air does not remember the original speed, and new air coming in does not keep its velocity, but settles down with the co-moving air. The speed it has is determined by the fan blowing it in, not by the speed of the train.
The reason is that the train pushes the air just as it pushes everything else. The air transmits the push by a pressure force, and there is no significant airflow inside the car when you start and stop, even at huge acceleration. Nothing is different from a stationary train, except during acceleration. The effect of acceleration will create a small pressure gradient in the air, and a density gradient, but these are insignificant, because the acceleration is slow.
This is counterintuitive to many people, but it is absolutely 100% true in the real world. Aristotle also confused things with air, despite the fact that Aristothenes, Archimedes, and other ancient scientists believed in some sort of inertia principle.and this type of thing
@JohnDuffield is indeed correct. Let me try to quantitatively show the anisotropy of the speed of light for an observer on the rotating platform. I think the following reasoning is to be credited to Langevin (I don't remember the reference, sorry, and anyway it would be in French!).
So let's denote by $\omega$ the angular speed of the platform with respect to the still observer. Clocks and rods on the rotating platform will be affected by their motion with respect to the still observer. What order in $\omega$ shall we expect? If there is a term proportional to $\omega$, then the clocks and rods will behave differently depending on whether the platform rotate in one or the other direction. This is silly, as this amount to simply having the still observer look from the top instead of from the bottom for example. So the first term has to be proportional to $\omega^2$. But then, that means that if we limit ourselves to an approximation at order $\omega$, we can completely neglect the fact that clocks and rods on the rotating platform will not measure time and lengths as for the still observer. What that means is that we can use a good old Galilean transform!
So let's denote by $(x, y, z, t)$ and $(x',y',z',t')$ the spacetime coordinates on the platform and of the still observer respectively, then
$$\begin{align}
x'&=x\cos\omega t-y\sin\omega t\\
y'&=x\sin\omega t + y\cos\omega t\\
z'&=z\\
t'&=t
\end{align}$$
In differential form, this reads
$$\begin{align}
dx'&=dx\cos\omega t-dy\sin\omega t - \omega (x\sin\omega t+y\cos\omega t)dt\\
dy'&=dx\sin\omega t + dy\cos\omega t + \omega(x\cos\omega t-y\sin\omega t)dt\\
dz'&=dz\\
dt'&=dt
\end{align}$$
Then we look at the spacetime interval $ds$. The still observer being inertial,
$$ds^2 = dt'^2-dx'^2-dy'^2-dz'^2.$$
I have used units where the speed of light $c=1$. Then substituting the above change of coordinates, and a bit of trigonometry, gives
$$ds^2 = dt^2 -2\omega(xdy-ydx)dt\underbrace{-dx^2-dy^2-dz^2}_{-dl^2}.$$
The propagation of a light signal corresponds to $ds^2=0$. The presence of cross terms $dxdt$ and $dydt$ results in an anisotropic speed of light. Let's investigate that precisely.
We note that $dA=\frac{1}{2}(xdy-ydx)$ is the area of the infinitesimal triangle whose vertices are the origin of the coordinates, the point $(x,y,z)$ and the point $(x+dx, y+dy, z+dz)$, that is to say the area swept by the vector from the origin to the position of the light signal during the duration $dt$. This is an area with a sign: positive if the sweeping is anti-clockwise and negative otherwise. So we get
$$dt^2 - 4\omega dAdt -dl^2=0.$$
We can solve this 2nd order equation in $dt$:
$$dt = 2\omega dA + \sqrt{dl^2+4(\omega dA)^2}.$$
But since we neglect terms of order $\omega^2$,
$$dt = dl + 2\omega dA,$$
i.e.
$$\frac{dl}{dt} = 1 - 2\omega \frac{dA}{dt}.$$
This is the speed of light for an observer on the platform: not only it is not equal to 1 but it depends on the direction of propagation since it depends on the sign of $dA/dt$.
Best Answer
First of all, weight which attached shouldn't be equal to floating force since rope and balloon have mass too. But let's say balloon is floating for us. Assuming that you have travelled with train at least once. You know that air molecules are not just back side of the train but everywhere. You can apply the same logic for floating balloon too.