# projectile power with air resistance [closed]

I need to get the x/y values of a fired projectile given the angle of the initial power, and air resistance.

The formulas I have are

$$V_x=V_o \cos\theta$$

$$X=V_o \cos \theta t$$

$$V_y=V_o \sin \theta -gt$$

$$Y=V_o \sin \theta t-(1/2)gt^2$$

To calculate the initial velocity $V_o$, I did

$$V_o= \sqrt{2E/m}$$

where

$E$ = power with which the ball was fired, $m$ = mass of ball

What I need is how to add air resistance to this equation

Besides, are my equations correct?

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## closed as too localized by David Z♦Apr 9 '11 at 23:29

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Note that Physics.SE uses to MathJax to provide LaTeX-like formatting of mathematical expressions. I've fixed up this post; you should learn to use it if you are going to participate on Physics.SE. –  dmckee Apr 9 '11 at 17:08
I think that your equations are not correct. Shouldn't $V_x,V_y$ be derivatives of $X,Y$? They surely aren't in your case. They're related by transformations that resemble the deletion of random characters. ;-) Moreover, $\theta$ seems to be dimensionless from some equations and frequency from other formulae. –  Luboš Motl Apr 9 '11 at 18:20
@Luboš, I interpreted Smith's equations as $X = (V_0 \cos\theta) t$ and $Y = (V_0 \sin\theta) t - (1/2)gt^2$ –  Keenan Pepper Apr 9 '11 at 18:56
What is this "Power" in headline or later last equation? It looks like energy! –  Georg Apr 9 '11 at 20:24
Try chapter 1 here: books.google.com/… –  Mark Eichenlaub Apr 9 '11 at 23:22

Often people model the drag in the simplest possible way, as a force proportional to the velocity but in the opposite direction: $\mathbf{F} = -b \mathbf{v}$ This is known as Stokes drag and is only approximately true at low velocities (low Reynolds numbers). At higher velocities the drag force becomes proportional to $v^2$ rather than $v$, and if the velocity ever approaches the speed of sound, the situation becomes more complicated with the possibility of shocks forming.