Drag Force of a Falling Object

## **Problem**

Consider a ball dropped form rest. The ball is subject to a drag force proportional to its velocity squared:

F

D

=
c

v

2

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

.

**Part 1:** Determine the terminal velocity.

**Hint**

The terminal velocity is the velocity in which the speed no longer accelerates whilst falling.

**Part 2:** Determine the velocity over time.

**Hint**

∫

0

x

d
x

a

2

−

x

2

=

1

2
a

ln
⁡

(

a
x

a
−
x

)

{\displaystyle \int \_{0}^{x}{\frac {dx}{a^{2}-x^{2}}}={\frac {1}{2a}}\ln {\left({\frac {a+x}{a-x}}\right)}}

**Part 3:** Determine the trajectory over time.

**Hint:**

1
−
x

1
x

=
1
−

2
x

1
x

{\displaystyle {\frac {1-x}{1+x}}=1-{\frac {2x}{1+x}}}

## **Solution**

**Part 1**

When the ball is released, the free-body diagram suggests that

F

n
e
t

=
−
m
g
c

v

2

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

.
The terminal velocity is the velocity in which the speed no longer accelerates whilst falling, so the net force must be zero.

# 0
−
m
g
c

v

t

2

{\displaystyle 0=-mg+c{v}\_{t}^{2}}

v

t

=

m
g

c

{\displaystyle {v}\_{t}={\sqrt {\frac {mg}{c}}}}

**Part 2**

To solve for the velocity over time, it is advantageous to transform the differential equation as follows. Remember the ball was originally at rest.

F

n
e
t

=
−
m
g
c

v

2

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

m
# a
−
m
g
c

v

2

{\displaystyle ma=-mg+cv^{2}}

m

d
v

d
t

=
−
m
g
c

v

2

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

d
v

d
t

=

−
g

v

t

2

(

v

t

2

−

v

2

)

{\displaystyle {\frac {dv}{dt}}={\frac {-g}{{v}\_{t}^{2}}}\left({v}\_{t}^{2}-v^{2}\right)}

Thus

∫

0

v

d
v

v

t

2

−

v

2

=

∫

0

t

−
g

v

t

2

d
t

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

and the solution is

1

2

v

t

ln
⁡

(

v

t

v

v

t

−
v

)

=

−
g

v

t

2

t

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

.
Solving for the velocity yields

v
(
t
# )
−

v

t

(

1
−
exp
⁡
(
−
2
g
t

/

v

t

)

1
exp
⁡
(
−
2
g
t

/

v

t

)

)

{\displaystyle v(t)=-{v}\_{t}\left({\frac {1-\exp(-2gt/{v}\_{t})}{1+\exp(-2gt/{v}\_{t})}}\right)}

.
**Part 3**

To determine the trajectory, use the fact

v
(
t
# )

d
x

d
t

{\displaystyle v(t)={\frac {dx}{dt}}}

.
Thus,

∫

x

0

x

d
# x

∫

0

t

−

v

t

(

1
−
exp
⁡
(
−
2
g
t

/

v

t

)

1
exp
⁡
(
−
2
g
t

/

v

t

)

)

d
t

{\displaystyle \int \_{x\_{0}}^{x}dx=\int \_{0}^{t}-{v}\_{t}\left({\frac {1-\exp(-2gt/{v}\_{t})}{1+\exp(-2gt/{v}\_{t})}}\right)dt}

.
The right-hand-side is difficult to integrate; however, it can be expressed as

∫

x

0

x

d
# x

∫

0

t

−

v

t

(

1
−

2
exp
⁡
(
−
2
g
t

/

v

t

)

1
exp
⁡
(
−
2
g
t

/

v

t

)

)

d
t

{\displaystyle \int \_{x\_{0}}^{x}dx=\int \_{0}^{t}-{v}\_{t}\left(1-{\frac {2\exp(-2gt/{v}\_{t})}{1+\exp(-2gt/{v}\_{t})}}\right)dt}

which makes it doable using integration by substitution.
The result is

x
(
t
# )

x

0

−

v

t

t
−

v

t

2

g

ln
⁡

(

1
exp
⁡
(
−
2
g
t

/

v

t

)

2

)

{\displaystyle x(t)=x\_{0}-{v}\_{t}t-{\frac {{v}\_{t}^{2}}{g}}\ln {\left({\frac {1+\exp(-2gt/{v}\_{t})}{2}}\right)}}

.