Projectile Motion with Stokes' Flow

Problem

Assume the air resistance is proportional to the projectile's velocity. Determine the trajectory of a projectile fired at origin, and with initial velocity

$$\begin{pmatrix} v_x \\ v_y \\ v_z \end{pmatrix}\ = \begin{pmatrix} \dot x_0 \\ \dot y_0 \\ \dot z_0 \end{pmatrix}\$$

and initial position

$$\begin{pmatrix} x \\ y \\ z \end{pmatrix}\ = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}\ .$$

Part 1: Determine the displacement vector.

Part 2: Let there be no motion in the y-axis. Determine the trajectory as the function $z(x)$.

Solution

Part 1

To model the forces of air resistance, consider the net force in three dimensions

$$\begin{pmatrix} ma_x \\ ma_y \\ ma_z \end{pmatrix}\ = \begin{pmatrix} -kv_x \\ -kv_y \\ -kv_z - mg \end{pmatrix}\ .$$

Divide the mass on both sides of the equation. Define $\rho = \frac{k}{m}$, and rewrite the vector quantities in terms of the coordinates $x, \; y, \;z .$

$$\begin{pmatrix} \ddot x \\ \ddot y \\ \ddot z \end{pmatrix}\ = \begin{pmatrix} -\rho \dot x \\ -\rho \dot y\\ -\rho \dot z - g \end{pmatrix}\$$

To obtain the velocity vector, integrate the acceleration with respect to time. Use the initial condition to solve for the constants.

$$\begin{pmatrix} \dot x \\ \dot y \\ \dot z \end{pmatrix}\ = \begin{pmatrix} \dot x_0 {e}^{-\rho t} \\ \dot y_0 {e}^{-\rho t} \\ \frac{-g}{\rho} + \left(\dot z_0 + \frac{g}{\rho} \right) {e}^{-\rho t} \end{pmatrix}\$$

To obtain the displacement vector, again, integrate the velocity with respect to time. Use the initial condition to solve for the constants.

$$\begin{pmatrix} x \\ y \\ z \end{pmatrix}\ = \begin{pmatrix} \frac{1}{\rho}\dot x_0 \left(1 - {e}^{-\rho t} \right) \\ \frac{1}{\rho}\dot y_0 \left(1 - {e}^{-\rho t} \right) \\ \frac{-g}{\rho} t + \frac{1}{\rho} \left(\dot z_0 + \frac{g}{\rho} \right) \left(1 - {e}^{-\rho t} \right) \end{pmatrix}\$$

Part 2

Since $x = \frac{1}{\rho}\dot x_0 \left(1 - {e}^{-\rho t} \right)$, it follows that

$$t = \frac{1}{\rho} \ln \left(\frac{\dot x_0}{\dot x_0 - \rho x} \right) .$$

Therefore, the function trajectory can be expressed as

$$z(x) = \frac{-g}{{\rho}^{2}}\ln \left(\frac{\dot x_0}{\dot x_0 - \rho x} \right) + \frac{x}{\dot x_0} \left(\dot z_0 + \frac{g}{\rho} \right)$$

or

$$z(x) = \frac{- m^2 g}{{k}^{2}}\ln \left(\frac{m\dot x_0}{\dot m x_0 - kx} \right) + \frac{x}{\dot x_0} \left(\dot z_0 + \frac{mg}{k} \right) .$$