This is not a question about throwing a ball vertically in a moving train. I am asking what would happen if I throw a ball in a horizontally in a moving train. Assume I am facing an exit door of a train and I throw a ball out of the exit door. Now my question is will the ball continue in a straight path for an outside observer or curve. How will it appear to me. I said curve because since the train is moving, the ball also has some initial sideways momentum perpendicular to the direction it is thrown in. The image below shows a visual description of the scene.

enter image description here

The train is moving at some velocity $v_x$ on the $z$-axis and the ball is thrown at some velocity $v_y$ along the x-axis. But the ball also has some initial velocity before it is thrown in the direction of the $z$-axis. So will the ball follow a curved trajectory along the $z$-axis?

The $z$-axis is the axis along which the train and the man are moving.


We neglect air friction and also the effects of gravity.


3 Answers 3


Let me first go through this without friction or air drag.

You say $v_y$ along the $x$-axis and the train moves with $v_x$ along the $z$-axis. This is a little inconsistent. I will use the velocities, but not your description of the axes. So the train moves in the $x$-direction, the ball is thrown into the $y$-direction and it the $z$-direction is up-down.

From the train

From the observer in the train, the ball will move with a constant $v_y$ away from the train. There is nothing that slows it down. Also, there is no $v_x$ component in the motion of the ball relative to the train. So the man on the train will see the ball right in front of him, flying further away and starting to fall down with $v_z = - g t$. There will be a curved trajectory, a parabola in the $y$-$z$-plane, the plane where the train is moving perpendicularly to.

That looks like this:

enter image description here

You can write this with vectors like so, with $g$ being the acceleration due to gravity: $$ \vec v(t) = v_y \, \hat y - g t \, \hat z = \begin{pmatrix} 0 \\ v_y \\ - gt \end{pmatrix} $$

Then you can integrate this again with respect to $t$ and get the position $\vec r$ of your ball. I set all integration constants to 0 to make this simpler. In principle, they allow for any starting point. I'll just assume that the starting point is the origin of the coordinate system. So the trajectory is: $$ \vec r(t) = v_y t \, \hat y - \frac12 g t^2 \, \hat z = \begin{pmatrix} 0 \\ v_y t \\ - 0.5 gt^2 \end{pmatrix} $$

From the ground

If you are an observer such that the train is moving with respect to you, you will see the ball moving with a constant velocity in $x$ and $y$, but also seeing it start to fall down. So you see a parabola in a plane that is crosswise to the axes.

I made another picture, you are looking at the front of the train, just a little skewed to see the 3D axes:

enter image description here

The velocities are similar, except that you have to include the motion of the train as well. The ball has the same $v_x$ as the train has. So this is $$ \vec v(t) = v_x \, \hat x + v_y \, \hat y - g t \, \hat z = \begin{pmatrix} v_x \\ v_y \\ - gt \end{pmatrix} $$

After integration, this is: $$ \vec r(t) = v_x t \, \hat x + v_y t \, \hat y - \frac12 g t^2 \, \hat z = \begin{pmatrix} v_x t \\ v_y t \\ - 0.5 gt^2 \end{pmatrix} $$

Galilei transformation

Alternatively, you could apply a Galilei transformation to it. I will try to be pedantic since transformations from coordinate systems are hard to get right. I just have done months of general relativity, so I know how hard this is :-)

Let the system of the train be system $\Sigma$ where the coordinates are $\vec r$ and $\vec v$. The system on the ground shall be $\tilde \Sigma$ where the coordinates are $\tilde{\vec r}$ and $\tilde{\vec v}$.

So we already had the following for the train (without $\tilde{}$): $$ \vec r(t) = v_y t \, \hat y - \frac12 g t^2 \, \hat z = \begin{pmatrix} 0 \\ v_y t \\ - 0.5 gt^2 \end{pmatrix} $$

Now the transformation from the train to the ground goes like the following: $v_x \to \tilde v_x = v_x + v_\text{Train}$. All other velocities are unchanged. When this is integrated, the space points will be transformed with $r_x \to \tilde r_x = r_x + v_\text{Train} t$.

With that transformation, we can get the trajectory viewed from $\tilde\Sigma$, the ground: $$ \tilde{\vec r}(t) = \underbrace{\begin{pmatrix} 0 \\ v_y t \\ - 0.5 gt^2 \end{pmatrix}}_{\vec r(t)} + \begin{pmatrix} v_\text{Train} t \\ 0 \\ 0 \end{pmatrix} $$

You said that the train was moving with $v_x$, so we can write $v_\text{Train} = v_x$ and get $$ \tilde{\vec r}(t) = \underbrace{\begin{pmatrix} 0 \\ v_y t \\ - 0.5 gt^2 \end{pmatrix}}_{\vec r(t)} + \begin{pmatrix} v_x t \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} v_x t \\ v_y t \\ - 0.5 gt^2 \end{pmatrix} $$ which we had earlier already.

Air drag

Air drag will cause the ball to slow down in each of its velocities, bending the curve even more.

Picture from above

If you look from above, this is the same as ignoring gravity. It looks like this:

enter image description here

Time snapshots

When you are in the train, you will see the rails moving below you, and the ball will just move in your $y$-direction:

enter image description here

When you are outside, you will see the train moving. The ball will always be in front of the person who threw it. Therefore it will move on a diagonal line. That line is straight, however!

enter image description here

  • $\begingroup$ Well I neglected gravitational effects to make it a 2D concept for simplification. Can you please tell me now will there be a trajectory in the z-axis of the ball? $\endgroup$ Jul 13, 2014 at 19:28
  • $\begingroup$ I added some math to show you how to calculate this in the coordinate systems and how to transform from one into the other. If you want to neglect the gravity, set $g = 0$. $\endgroup$ Jul 13, 2014 at 19:52
  • $\begingroup$ Well sorry even I am only 15 and do not understand math. Just answer my comment in yes or no. So the ball will bend w.r.t. x-axis? $\endgroup$ Jul 13, 2014 at 19:53
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    $\begingroup$ If you ignore air drag and gravity, the ball will go in a straight line! Newton's first postulate says that a body will remove in uniform motion (straight line) as long as no force acts on it. Since you excluded gravity and air drag, there are no forces left. If you look at the whole thing from the top, it will be a straight line. I'll add a picture. $\endgroup$ Jul 13, 2014 at 19:57
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    $\begingroup$ The 3D plots are made with Mathematica 9, the sketches are drawn in Xournal on a ThinkPad X220 Tablet with a Wacom digitizer. $\endgroup$ Jul 14, 2014 at 10:45

Assuming that the train is an inertial frame of reference (non-accelerating) and if we neglect both air friction and the effects of gravity, then the ball will move in a straight line away from you at the same speed which you threw the ball at until some other force acts on the ball. This situation would look the same as one where you threw the ball while standing still.

If someone was standing next to the train tracks at the spot where you released the ball, they too would see the ball moving in a straight line, they however would see the ball moving with the velocity in the z direction added to the velocity in the x direction. To understand how these vectors add, just look at this image below:

enter image description here

Vg is what the person floating next to the train would observe.

This question is illustrative of Newton's first law.

  • $\begingroup$ You can use LaTeX with $V_g$ to make the formula prettier. If the person floating next to the train observes $v_g$, it is at rest. However, it sounds like the person is floating (and moving) with the train. $\endgroup$ Jul 14, 2014 at 10:49

Let me make you precise about conventions, because your notations are unconventional. Suffix in a vector quantity is always represent the component along specific direction, component of velocity $v$ along x-axis is denoted by $ v_x.$ However, your question is precise irrespective of your conventions used.

Insight into your problem

You have made precise that ignore the effects of gravity.
Even if you throw the ball out of the train in perpendicular direction at some specific moment of time, it will not appear to be perpendicular to you when you observe from the train, But for an outside observer, it will be perpendicular to the train which is moving. Both for you and outside observer, the ball will take straight line as long as it is not accelerated.
Always have in mind, curve trajectory generally occurs in the condition where velocity and acceleration acting on a ball in perpendicular. In your question, it occurs in two cases

  1. when initial velocity imposed on body is subjected to acceleration in the perpendicular direction.i.e if you give $v_y$ along x-axis and there is acceleration acting, say $a_x$ along z-axis- as per your conventions, then there would be curved trajectory for both you and outside observer.

  2. when the ball is imposed with an acceleration $a_y$ along x-axis(as per your conventions), the ball would have curved trajectory for you alone, but for outside observer it would be in straight line path but will have acceleration.


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