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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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    @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 $y$.2012-01-11

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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 $