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.
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.
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 $\displaystyle t$)