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I am working on a series of equations to calculate the trajectory of a projectile in reverse. For example, given the ground impact angle and velocity, calculate the flight of the projectile. Here is the equation to calculate the drag on a projectile in normal flight (not reversed yet):

$$ x_2=x(1-k\sqrt{x^2+y^2}) $$ $$ y_2=y(1-k\sqrt{x^2+y^2}) $$

$x$ and $y$ are the before-drag velocities, and $k$ is a constant that includes the fluid density, reference area, drag coefficient, and timestep. The result, $x_2$ and $y_2$, are the after-drag velocities. On my graphing calculator, including these functions in a program creates a nice-looking trajectory. However, to reverse the formula, I need to solve the above equations for $x$ and $y$. Given the resulting velocity, find the velocity at the previous timestep. I am at a loss on how to do this.

Put simply, solve the above 2 equations for $x$ and $y$.

Any help will be appreciated.

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Could you please explain exactly what you're trying to do? Are the $x,y$'s infinitesmal quantities that are evaluated at each time step and then summed up (integrated) to calculate the trajectory? If so, what integration scheme are you using? – yohBS Jan 10 '12 at 20:47
@yohBS I don't understand exactly what you are saying (I'm in 10th grade). Every timestep, $x_2$ times the timestep is added to the x position of the projectile, and $y_2$ times the timestep is added to the y position. Then, $x$ and $y$ are set equal to $x_2$ and $y_2$ for the next timestep. Hope this helps. – Joel Jan 10 '12 at 21:11
You just do what you would do as if time was running forwards, but change the sign of the drag $k$ (is the opposite of drag shove?) and the two "after-drag velocities", so the projectile goes up and backwards with extra shove. – Henry Jan 10 '12 at 22:50
@Henry Changing the sign of $k$ doesn't work. That is applying negative drag to $x_2$ and $y_2$. For the equations to work, the drag must be applied to $x$ and $y$. – Joel Jan 11 '12 at 16:11

It helps to write down the equation in vector form: $$\mathbf v_2=(1-k|\mathbf v|)\mathbf v\tag1$$ This shows that $\mathbf v_2$ is a scalar multiple of $\mathbf v$ (no surprise there). Take the magnitude on both sides of(1): $$|\mathbf v_2|=(1-k|\mathbf v|)|\mathbf v|\tag2$$ Here I assume that $1-k|\mathbf v|\ge 0$, that is the velocity does not change direction due to drag. This is physically reasonable. Equation (2) is quadratic for $|\mathbf v|$, with the solution $$|\mathbf v|=\frac{1-\sqrt{1-4k|\mathbf v_2|}}{2k} \tag3$$ Thus, to obtain $\mathbf v$ from $\mathbf v_2$, we should divide $\mathbf v_2$ by its magnitude, and the multiply the result by the right side of (3): $$ \mathbf v =\frac{1-\sqrt{1-4k|\mathbf v_2|}}{2k |\mathbf v_2|}\mathbf v_2 $$

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