Throwing a ball from a train at an angle

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To negate the forward momentum of a train when throwing a ball at an angle, the ball's velocity must have two components: one opposing the train's motion and one perpendicular to it. The required speed can be calculated by dividing the train's speed by the cosine of the angle at which the ball is thrown. A scale diagram can also help visualize this, where the train's speed is represented as an arrow, and the ball's direction is drawn to find the necessary throwing speed. The discussion emphasizes the importance of understanding vector components in this scenario. Accurate calculations ensure the ball's motion effectively counteracts the train's speed.
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Hi all, I've been trying to find an answer to this question for a while, (bear with me physics is not a strong point). I would like to know how one would go about calculating how fast you have to throw a ball out of a train to negate its forward momentum. My understanding is that if you throw the ball straight out of the back you just need to match the trains speed, but what happens if you throw the ball at a diagonal? If i considered straight out of the side of the train to be 90degrees and out the back to be zero degrees, does the ball need to be thrown harder to compensate for the shallower angle and if so by how much?
 
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Yes you're quite right.

When you throw the ball at an angle, you can think of its motion as having two components - one in the direction the train is moving and one perpendicular to it, ie sideways.

And you want the component in the direction of the train to be exactly the opposite of the train's motion.

One way to see how fast this is for any direction, is to draw a scale diagram: draw an arrow backwards equal to the trains speed, then draw a line perpendicular until it meets the line of the ball's direction. Then the length of the line along the ball's direction tells you the speed you have to throw it. (I'll post a diagram as soon as I can do one.)

You can use a calculator and do some physics mumbo jumbo - just divide the speed of the train by the cosine of the angle you throw the ball and that's how fast you must throw it.
train.png
 
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Merlin3189 said:
Yes you're quite right.

When you throw the ball at an angle, you can think of its motion as having two components - one in the direction the train is moving and one perpendicular to it, ie sideways.

And you want the component in the direction of the train to be exactly the opposite of the train's motion.

One way to see how fast this is for any direction, is to draw a scale diagram: draw an arrow backwards equal to the trains speed, then draw a line perpendicular until it meets the line of the ball's direction. Then the length of the line along the ball's direction tells you the speed you have to throw it. (I'll post a diagram as soon as I can do one.)

You can use a calculator and do some physics mumbo jumbo - just divide the speed of the train by the cosine of the angle you throw the ball and that's how fast you must throw it.
View attachment 89081
Amazing, thanks.
 
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