Classical Relativity and the Speed of Light

[knowLittle]

Homework Statement

Let's assume that the classical ideas of space and time are correct, so that there could only be one frame, "ether", in which light traveled with same speed in all directions.

Assume that the Earth's speed relative to the ether frame is our orbital speed around the sun.

a.) What would be the observed speed (on earth) of a light wave traveling parallel to v?
b.) " " ... traveling anti-parallel to v?
c.) What if it were traveling perpendicular to v (as measured on earth)?

Homework Equations

$c = 2.9979 \times 10^8$ m/s
$v= 3\times10^4$ m/s

The Attempt at a Solution

Question C is my problem
Apparently, this is just...

^ c
|
| ----> v

However, I have seen that the hypotenuse is calculated as $\sqrt{c^2-v^2}$ it confuses me a bit.
I know that we are using vectors and v's direction would be ideally facing c, but I am still lost.

Or, are they assuming that :
^ c
|
| <----- v

In order to write the hypotenuse normally and then change the sign in v?
$\sqrt{c^2 + (-v^2)}= \sqrt{c^2-v^2}$ , but if this is the case, why not do something like this:

               ^ c
               |
               |
------> v

And, then write the hypotenuse without any change of sign?
Help?

 

[sonicharmony]

If you were traveling in a car going speed v, what would be the observed speed of a second card moving in the same direction in the lane next to you? What about if the other car was moving in the opposite direction in the lane next to you? For question C, think of your classic vector drawing. The vector AB has a component in the a direction, and a component in the b direction. AB is the hypotenuse of that triangle:

http://mathworld.wolfram.com/VectorAddition.html

So the Observed Speed V_{observed} = \sqrt{c^{2}+v^{2}}

It is only asking for speed (magnitude) not direction, so it seems to me it doesn't matter if the light is traveling perpendicular Up or perpendicular down, or if it is in front of or behind the path of the earth.