# Relativity: Rocket shoots a bullet straight down

1. Apr 19, 2012

### mitchmaniac20

1. The problem statement, all variables and given/known data

A rocket passes the earth at 0.8c. As it goes by, it launches a projectile at 0.6c perpendicular to the direction of motion. What is the projectile's speed in the earth's reference frame?

2. Relevant equations

[;u=\frac{u'+v}{1+u'v/{c}^{2}};]

3. The attempt at a solution

Since in the direction of the rocket's motion, u'=0, u=v=.8c. In the perpendicular direction since v=0, u'=u=.6c. However, this makes the overall speed of the projectile [;\sqrt{{.8c}^{2}+{.6c}^{2}}=c;], which shouldn't be possible.....

Last edited: Apr 19, 2012
2. Apr 19, 2012

### Steely Dan

It is incorrect to use the one-dimensional relativistic velocity addition formula if there are motions in multiple directions. In that case you need to take account of the more general vector form for velocity addition. This is essentially because even though all of the rocket's motion is perpendicular to the projectile's motion, time is still measured differently on the rocket than on Earth, so what the rocket sees as 0.6 c in the perpendicular direction, we will necessarily see as something different.

3. Apr 20, 2012

### mitchmaniac20

Thanks, this was actually part b where part a was to derive the velocity transformations in the direction perpendicular to motion, which I had incorrectly found to be u'=u, but your comment made me rethink that. Thanks for the help.