# Special Relativity Moving Frames Question

1. Jun 8, 2013

### knowLittle

1. The problem statement, all variables and given/known data
I am sort of reviving this thread.
My question is the same.
I understand and agree with what fisix responded a few years ago.
2. Relevant equations
cos( θ)= v_cart / v_bullet

3. The attempt at a solution
I can gather that in the reference frame of the cart, the bullet will appear to just go straight up and straight down.
And, in the reference frame of the ground, the bullet will look following a parabolic path.
Nevertheless, I cannot discern how to obtain the angle.

It seems to me that it has to be lesser or equal than pi/2.
I am not looking for the direct answer, but I just don't know how to proceed.

Thank you.

2. Jun 8, 2013

### ealbers

angle...

This reminds me of the Einstein thought experiment with a person standing on a moving train, tosses a ball straight up and sees it move straight up, then straight down right back into their hand.

Note, if you assume no air, the lack of a train surrounding the cart/cannon and ball makes no difference, you should shoot the cannon ball straight up, given the cart experiences no acceleration during the balls flight and there is no air resistance (as in a train, the air travels with you), the ball will naturally return from whence it was shot, the cart frame of reference is non-accelerating and therefore the ball remains in its frame at least in the direction of the carts travel, of course in the direction straight up it enters another frame, but at right angles to the carts direction and so the net effect is zero on that vector.

Now, from someone on the ground observing the cart, the cannon it carries and the ball it shoots, the situation appears very different.

Remember, the ball already has a velocity in the direction of the cart when it is 'tosses' up, and given no resistance from air, will happily keep this velocity during its flight.

So answer is, shoot the ball straight up! Don't accelerate the cart during flight, given no air resistance, get out of the way for its return as theres no terminal velocity involved to help slow the balls return :-)

Last edited: Jun 8, 2013
3. Jun 8, 2013

Thanks.