Will a ball thrown straight up in a train land in same spot (in real world)? I have a question that came up in a discussion with friends.  If I throw a ball straight up in an enclosed train car moving with constant velocity, I believe the basic physics books say it will land in the same spot.  But will it really?  I think I can say that the answer is "not in the real world".
Trivially, a train car is never enclosed.  Fresh air is being allowed into the carriage or the passengers would all die.  Thus there's currents of air that would affect the ball, agreed? If we remove the passengers and have a trusty robot (who does not need oxygen) throw the ball up in a carriage that really is completely air-tight, I'm still not sure it will land in the same spot.  I would imagine that there must still be air circulation.  The train had to start from a stop.  It's true the floor and the roof will drag the air right at the boundary along with it, but just as an open convertible car does not drag all the air in the world with it, I assume that the air in the middle of the car will not be dragged along at the same speed.  The air in the middle will remain stationary with respect to earth and pile up at the back of the car. Then it will be forced along.  I further imagine that this "pile of air" will try to redistribute itself uniformally. Won't all this set up currents?  Will the air come to be completely still in the reference frame of the car? [I'm guessing the answer is yet] How long would this take?  
Bonus question:  I believe if I'm sitting in a convertible car and throw a ball straight up it will land back in my hand as long as I don't throw it too far up.  At some point, I'll throw it to high and will lose the ball out the back of the car.  What's the relevant equation covering this in a car travelling at X miles per hour in still air?  Put another way, I'm trying to get a feel for how extensive the "boundary" layer of air around the car is and how it dissipates with distance.
 A: 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 
A: No, the ball does not land in exactly the same spot, even in the absence of any air at all.  However, in the real world, the effects in the train are probably too small to observe without a careful experimental setup.
dmckee's answer describes physics in an inertial reference frame.  However, because the Earth is spinning, the ball experiences a velocity-dependent Coriolis force.  This force comes from the ball's motion and the Earth's rotation, so its magnitude is on the order of $mv/T$, with $v$ the ball's velocity, $m$ its mass, and $T$ the time it takes Earth to rotate - one day.  The exact magnitude depends on the angle between the train's velocity and Earth's axis.  The force is perpendicular to the ball's velocity.
If this force pushes the ball to the side, off its trajectory, the distance it goes off should be on the order of $at^2$ with $a$ the acceleration and $t$ the time it's in the air.  If it's in the air for about a second, the deflection is small because a second is small compared to a day.  You'll get something around $10^{-5}v\textrm{s}$, so even if the train is going $100 m/s$ you still only get millimeters of deflection.
If the train is traveling at constant velocity, any oscillations induced a while ago by accelerating should equilibrate, and the air should be like normal still air (except again for Coriolis forces on the air).  As Ron Maimon mentioned, this happens pretty fast.  How fast?  Try singing in the shower.  When you hit a good resonance, suddenly stop singing and see how long the note keeps sounding.  Maybe a tenth of  a second.  Physically, that's roughly the same mechanism, so the oscillations of air set up by the acceleration of the train will die out on a similar timescale, plus or minus an order of magnitude to adjust for the size of the car and the boundary conditions at the walls.  
How about bulk motion of the air?  Try turning on a fan, then suddenly stopping it with a stick?  How long do you still feel wind?  Maybe a little longer, on the order of seconds this time, but experience still informs us that this dies off quickly, too.  I think Ron's point in comments that there essentially are no oscillations or currents set up by the train's acceleration is right.  The time scale on which the air equilibrates is fast compared to the time over which the train accelerates, so the air is essentially always in equilibrium.
A: I'll discuss two situation:


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*The experimenter is standing on a "X" in the enclosed railcar as it rolls along with velocity $v$, and throws just as she passes a "Y" painted on the ground outside the train. In this case she throws the ball "straight up" in her own frame.
She predicts that the ball lands on the "X", and is not disappointed.
A pedestrian watching from the verge also predicts that the ball lands on the "X", but in his frame of reference, this will be some distance $v\Delta t$ along the tracks (thus not on the "Y") when it does that.

*The experimenter carefully (very carefully) throws the ball such that the pedestrian see it go "straight up". This will require that she throws it toward the back of the train. 
Unsurprisingly she predict that it will not land on the "X", but back along the train by a distance $v \Delta t$.
The watcher predicts correctly that the ball will land beside the "Y", which it does.
In neither case do the very modest air currents in the car make an appreciable difference.
In the event that the railcar is open to the air, the we would need to make a fairly complicated calculation involving wind resistance, but it is still true that both observes will make the same predictions.
A: Even without regarding air issues, I think in a 'real world' scenario the ball will never drop at exactly the same spot because of minor accelerations during the ride.
When a car / train moves, I don't think it ever moves at a completely steady velocity (in fact, although I can't explain it, I have a feeling it's physically impossible).  There will always be minor accelerations of some sort.
A: I think this thread got spammed and floated to the top, but it reminded me of a similar demonstrations about frames of velocity.  Watch this video where a ball is fired from a moving vehicle yet if conditions are just right it falls straight down.  Ball drops straight down when fired from cannon
