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## Homework Statement

(d) A spaceship moves with velocity (4/5)c relative to Earth. The astronaut throws his empty beer bottle out the window with velocity (3/5)c relative to the ship in the sideways direction. What is the velocity of the bottle (magnitude and direction) relative to the Earth (i) classically, (ii) relativistically?

## Homework Equations

Velocity Addition (relativistic)

[tex]u_x = \frac {u_x' + v}{1 + \frac{v}{c} \frac{u_x'}{c}} [/tex]

[tex]u_y = \frac {u_y' + v}{\gamma (1 + \frac{v}{c} \frac{u_x'}{c})} [/tex]

Pythagorean Theorem

## The Attempt at a Solution

I just want to know if I've done this correctly. This is a problem on a practice exam.

For the classical portion I just used the Pythagorean theorem [tex] \sqrt {(\frac{4c}{5})^2 + (\frac{3c}{5})^2} = c [/tex]

For the relativistic portion I use the Velocity Addition equations. I use [tex]v = \frac{4}{5}c = u_x' [/tex]and [tex]u_y' = \frac{3}{5}c [/tex]

This gives:

[tex]u_y = \frac{15}{41}c [/tex]

and using the Pythagorean theorem with u_x = (4/5)c I get [tex] \frac{13}{15}c = .866 c [/tex]

I am embarrassed to say that I only think it's the right answer because it looks like it would be. Conceptually I have no idea why I would set both u_x' and v equal to 4/5 c. Can anybody explain if I did get it correct, and if I didn't how I might get the correct answer? Thank you!