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.

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- Nov 13th 2012, 02:06 PMjeremy5561Simplify 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 PMchristianwosRe: Simplify this function
Well, you can put the quantity under the square root under common denominator v

_{0}so that you will cancel the v_{0}at the outside and will get [v_{0}^{2}-2gtv_{0}sinx+(gt)^{2}]^{1/2} - Nov 14th 2012, 05:02 PMskeeterRe: Simplify this function
- Nov 14th 2012, 05:08 PMjeremy5561Re: 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 PMskeeterRe: Simplify this function
fix your equations ...

$\displaystyle v_x = v_0 \cos{\theta}$

(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 $\displaystyle t$) - Nov 15th 2012, 12:44 PMjeremy5561Re: Simplify this function
sorry, am really tired. I did it correctly originally. Don't think it can be simplified further.