A ball is thrown upward in a train moving with a constant velocity. Where will it land?

My intuition tells me that the ball will fall at my back. But my book says that it will return back to the thrower.

  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/7479/2451 , physics.stackexchange.com/q/14993/2451 , and links therein. $\endgroup$
    – Qmechanic
    Apr 12 '14 at 16:27
  • $\begingroup$ If the thrower is on the train than he is unknowingly also moving at the same speed as the train. Thus the ball that he throws will have an horizontal velocity equal to the speed of the train. It is this horizontal speed that makes the ball return in the hands of the thrower. If you were to observe the scene from the platform, you would see that the train, the passenger and the ball all move forward: the ball also has an upward velocity so it will describe a parabolic trajectory. $\endgroup$
    – SuperCiocia
    Apr 12 '14 at 16:34
  • $\begingroup$ @Harold I forgot to notice before I asked the question, the fact that there exists something called Newton's First Law Of Motion... $\endgroup$ Apr 12 '14 at 16:42
  • $\begingroup$ well yes that is THE reason, I was trying to give you an intuitive picture $\endgroup$
    – SuperCiocia
    Apr 12 '14 at 17:49
  • $\begingroup$ 1) Do you mean straight upward; and if yes, 2) In whose frame of reference is the ball thrown straight up $\endgroup$
    – DJohnM
    Apr 12 '14 at 17:59

Since the gravitational force only pulls the ball down, but not back or forth, it will not experience any acceleration changing its forward velocity but only downward acceleration. Thus, the ball will return to the thrower.

You can also imagine the train to have no windows and be moving extremely smoothly. The thrower won't know if the train is moving or not and so the ball won't "know" either. The scenario for the train at rest is quite intuitive, I think, and will not differ from the scenario of the train moving at constant velocity.

  • $\begingroup$ Clear and understandable +1!! $\endgroup$ Apr 12 '14 at 15:39

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