1. ## 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.

2. ## 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

3. ## Re: Simplify this function

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?

4. ## 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.

5. ## 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 $\displaystyle t$)

6. ## Re: Simplify this function

sorry, am really tired. I did it correctly originally. Don't think it can be simplified further.