**Free Fall with air resistance near earth's surface**

Approximations: NEAR earth's surface means we will use F_{gravity}
= -mg where g = constant; also we will assume that the air density is constant
and assume that the air resistance coefficient is indeed a constant.

**Case 1: F _{AR} = - bv**

**SF = ma** leads to **-mg - bv =
m dv/dt** , or

**dt**** =**** dv / [-g - (b/m)v] =
-(m/b) dv / [v + mg/b]**

Let **w = v + mg/b** so that **dw**** = dv**, and remember to adjust the
limits of integration:

**t = -(m/b) _{wo=vo+mg/b}**

**e ^{-bt/m} =
(v+mg/b) / (v_{o}+mg/b)**

**(v+mg/b) = (v _{o}+mg/b)
e^{-bt/m}**

**v(t) = v _{o}
e^{-bt/m} - (mg/b)(1- e^{-bt/m})**

*Units:
*1) (mg/b) should have units of velocity:
m has units of kg; g has units of m/s

*Limiting cases:
*1) as t goes to zero, v goes to v

It is unclear what happens for small time with small b since b is in the denominator of the coefficient! We can find out by expanding

v(small t)
= - (mg/b)(1 - [1 - bt/m + (1/2)b^{2}t^{2}/m^{2} -
(1/6)(b^{3}t^{3}/m^{3}) + ...]

**v(****small t) = -gt + (1/2)bgt ^{2}/m - (1/6)b^{2}gt^{3}/m^{2}
+ ...**

Thus, as b goes to zero, the speed becomes that of free fall without air resistance (v = -gt) like it should.

**Method of Successive Approximation Technique**

To put in effects that are smaller (such as air resistance compared to gravity),
we can use a method called the method of successive approximations.

In this method, we first solve the problem without any of the smaller effects:
(we will use v_{o}=0 to make the math easier)

Zero order - assume for now that b=0, and solve for v^{(}^{0)}(t)
:

**F = -mg** so **SF = ma** leads to: **-mg = m dv/dt** which gives: **dv = -g dt** , and integrating gives: **v ^{(0)}(t) = v_{o} - gt = -gt** .

First order use zero order solution for v in the air resistance term of -bv, and solve for v^{(}^{1)}(t):

**F = -mg - bv ^{(0)}** so

Second order use first order solution for v in the air resistance term of
bv, and solve for v^{(}^{2)}(t):

**F = -mg - bv ^{(1)}** so

Note that this gives the same result as the series expansion of our analytical result!

**-------------------------------------**

**Case 2: F _{AR} = - bv^{2} **Note that since v is squared, we need to treat the sign for F

For now, lets consider the case going downward, and choose down as the positive direction.

**SF = ma** leads to **mg bv ^{2} = m
dv/dt** , or

Let (b/mg)^{1}^{/2}v = sin(j) so that dv = (mg/b)^{1/2}cos(j)dj , and
remember to adjust the limits of integration: j = sin^{-1}[(b/mg)^{1/2}v]

**t = (1/g) **_{jo}**ς _{ }**

From a table of integrals, we have ς dj / cos(j) = ln [(1/cos(j) + tan(j)] so

**t =
(m/gb) ^{1/2} { ln [(1/cos(**

where j = sin^{-1}[(b/mg)^{1/2}v] and j_{o}
= sin^{-1}[(b/mg)^{1/2}v_{o}] .

Note: this is just t(v), it is not v(t). Getting from t(v) to v(t) looks quite imposing.

*Check units:
*1) (m/gb)

*Limiting cases:
*1) as v goes to v

** Homework Problem:** Problem #10. Use the method of successive approximation to
find the first order expression for v(t) for the above
Case 2.

__Alternate solution for case 2:__

**SF = ma** leads to **mg bv ^{2} = m
dv/dt** , or

Let (b/mg)^{1}^{/2}v = u so that dv = (mg/b)^{1/2}du , and
remember to adjust the limits of integration.
Note that (mg/b)^{1}^{/2} = v_{t} (see above), so (b/mg)^{1/2} = 1/v_{t}
, so we can see that u = v/v_{t} which is
dimensionless as it should be if we are to subtract its square from 1.

**t =**** (1/g) _{uo}**

To solve this integral, we note that 1/(1-u^{2}) =
A/(1-u) + B/(1+u) = {A(1+u) + B(1-u)}/(1-u^{2})

so that A+B = 1 and
A-B = 0 which has the solution of
A = ½ = B; so 1/(1-u^{2}) = ½ {1/(1+u)} + ½ {1/(1-u)}, so

**ς **du/(1-u^{2})
= ½ **ς **du/(1+u) + ½ **ς **du/(1-u)
;

now let y = 1-u and z = 1+u so that

**ς **du/(1-u^{2})
= ½ **ς **dz/z + ½ **ς -**dy/y = ½
{ln(z) ln(z_{o})} - ½ {ln(y) ln(y_{o})}

but since z_{o} = 1+u_{o} = 1+0 = 1,
and y_{o} = 1-u_{o} = 1, and ln(1) =
0, we have

**ς **du/(1-u^{2}) = ½ {
ln(1+u) ln(1-u) } = ½ ln{(1+u)/(1-u)} =

½ ln[ {1+(b/mg)^{1/2}v} / {1-(b/mg)^{1/2}v} ], so that

**t =**** ½ (m/gb) ^{1/2}
ln[ {1+(b/mg)^{1/2}v} / {1-(b/mg)^{1/2}v} ].**

Although the two expressions for t,

**t = (m/gb) ^{1/2}
{ ln [{(1/cos(**

**t =**** ½ (m/gb) ^{1/2}
ln[ {1+(b/mg)^{1/2}v} / {1-(b/mg)^{1/2}v} ] **

look quite different, they are in fact equivalent.

The second form we can convert from t(v) into v(t):

**2t(gb/m) ^{1/2} = ln[
{1+(b/mg)^{1/2}v} / {1-(b/mg)^{1/2}v} ]**

now let **a =** **(gb/m) ^{1/2} **and
we get:

**e ^{2}**

**e ^{2}**

**(b/mg) ^{1}^{/2}v (e^{2}**

**v(****t)
= (mg/b) ^{1/2} (e^{2}**

**v(****t)
= (mg/b) ^{1/2} (e**

**v(t) = (mg/b) ^{1/2} tanh[(gb/m)^{1/2} t] ** =

*Units:*

1. (mg/b)^{1/2} should have
units of m/s: from F=-bv^{2}, b
has units of Nt/(m/s)^{2} = {kg*m/s^{2}}/{m^{2}/s^{2}}
= kg/m; so (mg/b) = (kg)*(m/s^{2})/(kg/m) = m^{2}/s^{2}
and when we take the square root gives m/s.

2. (gb/m)^{1}^{/2}*t should be dimensionless: [(m/s^{2})*(kg/m)/kg]^{1/2}*s
= dimensionless as required.

*Limiting cases:*

For small t, which really means [(gb/m)^{1/2} t]
= gt/v_{t}_{ } is small, we need to look at the behavior of tanh(x) = (e^{x} e^{-x}) / (e^{x}
+ e^{-x}) by using the ^{x} and e^{-x} .

(e^{x} e^{-x})
/ (e^{x} + e^{-x}) ≈ ({1+x)} {1-x}) / ({1+x)} + {1-x}) = 2x
/ 2 =
x (to first order)

So for small t, (i.e., gt << v_{t}), we
have approximately:

v(small t) = **(**mg/b)^{1/2}
tanh[(gb/m)^{1/2}
t] =
v_{t} tanh[gt/v_{t}]** ** ≈
v_{t} * [(gt/v_{t}]** ** = gt , which is the case for no air resistance.

For small t to the next order in x using Taylor series for e^{x}, e^{-x},
and 1/(1+x):

e^{x} ≈ 1 + x +
x^{2}/2 + x^{3}/6
and e^{-x} ≈ 1 - x
+ x^{2}/2 - x^{3}/6 so
that

tanh(x) ≈ [(1 + x + x^{2}/2 + x^{3}/6 ) (1 - x + x^{2}/2
- x^{3}/6)] / [(1 + x + x^{2}/2 + x^{3}/6 ) + (1 - x +
x^{2}/2 - x^{3}/6)]

Now we use the approximation 1/(1+y) ≈
(1-y) to get

tanh(x) ≈ (2x + 2x^{3}/6) / (2 + 2x^{2}/2)
≈ (x + x^{3}/6) * (1 - x^{2}/2) ≈
x + x^{3}/6 x^{3}/2
= x x^{3}/3 so that
(recalling that v_{t}^{2} = mg/b)

v(t) = v_{t} tanh(gt/v_{t}) ≈ v_{t} [gt/v_{t}
g^{3}t^{3}/(3v_{t}^{3})] = gt
bg^{2}t^{3}/(3m) (where
down is positive).

Compare this approximation to the one in Problem 10 above.

For large t, tanh(x) = (e^{x}
e^{-x}) / (e^{x} + e^{-x}) goes to 1, and so v(large
t) goes to (mg/b)^{1/2} = v_{t} .

__Estimation of a real case:__

What is the terminal speed for an ice pellet (hail) if the pellet is a sphere of radius 1 cm?

If we work from the bv^{2} model for air resistance, and use b =
(1/2)C_{D}Aρ [see the earlier section on
F(v)], and knowing the density (ρ=m/V) of water (and ice) is close to 1
gm/cm^{3} = 1000 kg/m^{3} and air is close to 1.3 kg/m^{3},
and the formulas for a sphere: cross
sectional area (A) is that of a circle = pr^{2} and the
volume (V) is (4/3)pr^{3},
and using C_{D} = ½ for a sphere, we get:

m = ρ*V = (1000 kg/m^{3})*(4/3)p(.01 m)^{3} = .0042 kg = 4.2 grams;

A = pr^{2} = p (.01 m)^{2} = 3.1 x 10^{-4} m^{2};

b = (1/2)C_{D}Aρ_{air} = (1/2)(1/2)(3.1 x 10^{-4}m^{2})(1.3
kg/m^{3}) = 1.02 x 10^{-4} kg/m ;

v_{t} = [mg/b]^{1/2} = [(.0042
kg)(9.3 m/s^{2})/(1.02 x 10^{-4 }kg/m)]^{1/2} = 20.1
m/s = 45 mph.

If we increase the radius of the ice sphere by 10 (from 1 cm to 10 cm), m
goes up by a factor of 1000, A goes up by a factor of 100, b goes up by a
factor of 100, and v_{t} goes up by a factor
of (10)^{1/2} = 3.16, so v_{t} goes
up to 142 mph.