# Solve Drag Equation for Previous Timestep

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