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

(

v
x

v
y

v
z

)

=

(

x
˙

0

y
˙

0

z
˙

0

)

{\displaystyle \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

(

x

y

z

)

=

(

0

0

0

)

.

{\displaystyle \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
)

{\displaystyle z(x)}

.

## **Solution**

**Part 1**

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

(

m

a
x

m

a
y

m

a
z

)

=

(

−
k

v
x

−
k

v
y

−
k

v
z

−
m
g

)

.

{\displaystyle \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

# ρ

k
m

{\displaystyle \rho = \frac{k}{m} }

, and rewrite the vector quantities in terms of the coordinates

x
,

y
,

z
.

{\displaystyle x, \; y, \;z . }

(

x
¨

y
¨

z
¨

)

=

(

−
ρ

x
˙

−
ρ

y
˙

−
ρ

z
˙

−
g

)

{\displaystyle \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.

(

x
˙

y
˙

z
˙

)

=

(

x
˙

0

e

−
ρ
t

y
˙

0

e

−
ρ
t

−
g

ρ

(

z
˙

0

g
ρ

)

e

−
ρ
t

)

{\displaystyle \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.

(

x

y

z

)

=

(

1
ρ

x
˙

0

(

1
−

e

−
ρ
t

)

1
ρ

y
˙

0

(

1
−

e

−
ρ
t

)

−
g

ρ

t

1
ρ

(

z
˙

0

g
ρ

)

(

1
−

e

−
ρ
t

)

)

{\displaystyle \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

1
ρ

x
˙

0

(

1
−

e

−
ρ
t

)

{\displaystyle x = \frac{1}{\rho}\dot x\_0 \left(1 - {e}^{-\rho t} \right) }

, it follows that

# t

1
ρ

ln
⁡

(

x
˙

0

x
˙

0

−
ρ
x

)

.

{\displaystyle 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
# )

−
g

ρ

2

ln
⁡

(

x
˙

0

x
˙

0

−
ρ
x

)

x

x
˙

0

(

z
˙

0

g
ρ

)

{\displaystyle 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
# )

−

m
2

g

k

2

ln
⁡

(

m

x
˙

0

m
˙

x
0

−
k
x

)

x

x
˙

0

(

z
˙

0

m
g

k

)

.

{\displaystyle 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) .
}