Stone dropped from a moving train 
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*This may look like a stupid question, but it is really getting to me. Imagine a train moving with an acceleration $a$, and a person drops a stone from the window. To an observer on the ground, the stone follows a parabolic path, as it is a projectile with initial velocity the same as the velocity of train when dropped. However, the person who dropped it sees that it falls down in a straight line. Why? Can someone explain the reason to me?

*What will the acceleration of the stone measured by an observer on the ground. I think it should be $a_{net}=\sqrt{g^2 + a^2}$, but I have a book which says it should be $g$. No explanation is provided. Help me out please.
 A: *

*You're not correct that the stone will appear to fall straight down if the train's acceleration is $\neq 0$. It would appear to fall in a (straight) slanted line (with angle $\arctan\left(\frac{a}{g}\right)$) because the stone is accelerating in the $y$ axis (due to gravity) and the observer in the train is accelerating in the $x$ axis. From outside the train, it would appear to fall parabolically because it has an initial velocity in the $x$ axis (due to the fact it was moving with the train when it was dropped) and it accelerates in the $y$ axis.

*The answer is $g$ because the only force acting on the stone is gravity, and the acceleration on all free falling bodies on earth is $g$.
It might help if you draw a free body diagram of the problem to see why this is the case. It looks like you're a bit confused about why things accelerate. Acceleration is always due to a force, and if you can't point to a force that causes the acceleration (like: gravity, friction on the rails moving the train forward, etc) then there can't be acceleration in that direction. What forces are acting on the stone that make you think it'll keep accelerating with the train once it's separated from the train?
A: 
To an observer on the ground, the stone follows a parabolic path, as it is a projectile with initial velocity the same as the velocity of train when dropped. However, the person who dropped it sees that it falls down in a straight line.

The stationary observer himself being at rest observes that the stone has a horizontal velocity and a vertical downward acceleration. So to him the stone has a parabolic path.But the man in the train will actually see the stone fall in a straight slanted line. Because the man in the train has an additional acceleration of 'a' along the horizontal direction, which the stone does not have. So he sees the stone with a relative horizontal deceleration and a downward acceleration. Hence he sees the stone fall in a 'straight slanted line' and not in a straight line as the train is accelerating**.

What will the acceleration of the stone measured by an observer on the ground. I think it should be $a_{net}=(g^2+a^2)^{1/2}$, but I have a book which says it should be g.

Here, I must say that when a stone is dropped from an accelerating train, the moment the stone loses contact with the man on the train, it no longer experiences the horizontal acceleration. It has a free fall ,but it retains the horizontal velocity it had at the moment the stone loses contact with the man on the train. Hence the net acceleration is 'g'.
A: The path of the stone for the observer on the train should be a straight line as, when releasing the stone, it had the velocity same as train. However, air resistance may affect the stationary path. For the observer on ground it should be parabolic due to the action of gravity on the moving stone. The net acceleration of stone should be $g$ since after dropping it from the train it is only accelerated by gravity. 
A: First, shall I mention that you have not mentioned air resistance in your problem. Because in this example, air resistance would mean that the observer on the train will actually not see the stone fall in a straight line (including the diagonal mentioned by @Aniket) and there is a horizontal deceleration experienced. 
