# Simplify this function

• Nov 13th 2012, 02:06 PM
jeremy5561
Simplify this function
I have an equation that looks like this
$\displaystyle v=v_0 sqrt(1-2gt sin\theta /v_0+(gt/v_0)^2)$

Is there any easy way to simplify this? It's almost a perfect square.
• Nov 14th 2012, 04:52 PM
christianwos
Re: Simplify this function
Well, you can put the quantity under the square root under common denominator v0 so that you will cancel the v0 at the outside and will get [v02-2gtv0sinx+(gt)2]1/2
• Nov 14th 2012, 05:02 PM
skeeter
Re: Simplify this function
Quote:

Originally Posted by jeremy5561
I have an equation that looks like this
$\displaystyle v=v_0 sqrt(1-2gt sin\theta /v_0+(gt/v_0)^2)$

Is there any easy way to simplify this? It's almost a perfect square.

have to ask ... is there more to this than meets the eye? in other words, how did this equation come about?
• Nov 14th 2012, 05:08 PM
jeremy5561
Re: Simplify this function
For a projectile launched with velocity $\displaystyle v_0$ at angle $\displaystyle \theta$, what is the magnitude of the velocity vector at time $\displaystyle t$?

$\displaystyle v_0_x=v_0cos\theta t$
$\displaystyle v_0_y=v_0sin\theta t-gt$

Apply the pythagorean theorem and you'll end up where I am.

Unless I did something wrong. Stayed up till 2 AM doing this stuff.
• Nov 14th 2012, 05:14 PM
skeeter
Re: Simplify this function
$\displaystyle v_x = v_0 \cos{\theta}$
(note there is no $\displaystyle t$ ... velocity in the x-direction remains constant)
$\displaystyle v_y = v_0 \sin{\theta} - gt$
(first term does not have a $\displaystyle t$)