Skydiver

## **Problem**

   Consider a skydiver falling out of an airplane. He opens his parachute during the descent. Approximate the external force on the skydiver, given that air resistance is proportional to the velocity squared

F

d
r
a
g

=
c

v
2

{\displaystyle {F}\_{drag} = cv^2 }

.

**Part 1:** Use the force equation to calculate the terminal velocity

v

τ

{\displaystyle {v}\_{\tau} }

.

**Part 2:** Change your force equation into a first-order differential equation. Then use your result to arrive at the integral

∫

0

v

d
v

v

τ

2

−

v
2

=
−

g

v

τ

2

∫

0

t

d
t
.

{\displaystyle \int\_{0}^{v} \frac{dv}{v\_{\tau}^{2} - v^2} = -\frac{g}{v\_{\tau}^{2}} \int\_{0}^{t} dt. }

**Part 3:** Solve the above integral. Then rearrange the velocity solution as a function of time.

## **Hint**

(1) Use the integral

∫

1

a
2

−

x
2

d
# x

1

2
a

ln
⁡

(

a
x

a
−
x

)

C
.

{\displaystyle \int \frac{1}{a^2 - x^2} dx = \frac{1}{2a} \ln{\left(\frac{a + x}{a - x} \right)} + C. }

## **Solution**

**Part 1**

There are two forces on the skydiver: the force of gravity and the force of air-resistance (drag force):

F

n
e
t

=

F
g

F

d
r
a
g

.

{\displaystyle {F}\_{net} = F\_g + {F}\_{drag}. }

Since

F
g

=
−
m
g

{\displaystyle F\_g = -mg }

 and

F

d
r
a
g

=
c

v
2

{\displaystyle {F}\_{drag} = cv^2 }

, we get

F

n
e
t

=
−
m
g
c

v
2

.

{\displaystyle {F}\_{net} = -mg + cv^2. }

The falling object is in equilibrium when it reaches terminal velocity, therefore,

# 0
−
m
g
c

v

τ

2

{\displaystyle 0 = -mg + cv\_{\tau}^{2} }

c

v

τ

2

=
m
g

{\displaystyle cv\_{\tau}^{2} = mg }

v

τ

=

m
g

c

.

{\displaystyle v\_{\tau} = \sqrt{\frac{mg}{c}}. }

**Part 2**

Newton's second law of motion states

F

n
e
t

=
m
a

{\displaystyle F\_{net} = ma }

. Since

# a

d
v

d
t

{\displaystyle a=\frac{dv}{dt}}

, we get

m

d
v

d
t

=
−
m
g
c

v
2

.

{\displaystyle m\frac{dv}{dt} = -mg + cv^2. }

Since

# c

m
g

v

τ

2

{\displaystyle c = \frac{mg}{v\_{\tau}^{2}} }

,

d
v

d
t

=
−
g

g

v

τ

2

v
2

{\displaystyle \frac{dv}{dt} = -g + \frac{g}{v\_{\tau}^{2}}v^2 }

v

τ

2

d
v

d
t

=
−
g
(

v

τ

2

−

v
2

)
.

{\displaystyle v\_{\tau}^{2}\frac{dv}{dt} = -g(v\_{\tau}^{2} - v^2). }

Use separation of variables to arrive at the integral

∫

0

v

d
v

v

τ

2

−

v
2

=
−

g

v

τ

2

∫

0

t

d
t
.

{\displaystyle \int\_{0}^{v} \frac{dv}{v\_{\tau}^{2} - v^2} = -\frac{g}{v\_{\tau}^{2}} \int\_{0}^{t} dt. }

**Part 3**

Integrate using partial fractions

1

2

v

τ

ln
⁡

(

v

τ

v

v

τ

−
v

)

=
−

g

v

τ

2

t
.

{\displaystyle \frac{1}{2{v}\_{\tau}}\ln{\left(\frac{{v}\_{\tau} + v}{{v}\_{\tau} - v} \right)} = -\frac{g}{v\_{\tau}^{2}} t. }

Rearrange the above equation

(

v

τ

v

v

τ

−
v

)

=

e

−

2
g

v

τ

t

{\displaystyle \left(\frac{{v}\_{\tau} + v}{{v}\_{\tau} - v} \right) = {e}^{-\frac{2g}{{v}\_{\tau}} t} }

v

τ

# v

e

−

2
g

v

τ

t

v

τ

−

e

−

2
g

v

τ

t

v

{\displaystyle {v}\_{\tau} + v = {e}^{-\frac{2g}{{v}\_{\tau}} t} {v}\_{\tau} - {e}^{-\frac{2g}{{v}\_{\tau}} t} v }

v

e

−

2
g

v

τ

t

# v
−

v

τ

e

−

2
g

v

τ

t

v

τ

{\displaystyle v + {e}^{-\frac{2g}{{v}\_{\tau}} t} v = -{v}\_{\tau} + {e}^{-\frac{2g}{{v}\_{\tau}} t} {v}\_{\tau} }

v
(
t
# )
−

v

τ

(

1
−

e

−

2
g

v

τ

t

1

e

−

2
g

v

τ

t

)

.

{\displaystyle v(t) = -{v}\_{\tau} \left(\frac{1-{e}^{-\frac{2g}{{v}\_{\tau}} t}}{1+{e}^{-\frac{2g}{{v}\_{\tau}} t}} \right). }