Free Fall with air resistance near earth's surface EXTRA
Approximations: NEAR earth's surface means we will use Fgravity = -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 2: Falling down with FAR = - bv2 (extended)
Note that since v is squared, we need to treat the sign for FAR differently going up versus going
down. For now, let’s consider the case going downward, and choose down as the positive
direction.
F = ma leads to mg – bv2 = m dv/dt , which gives (see regular page on free fall):
t = ½ (m/gb)1/2 ln[ {1+(b/mg)1/2v} / {1-(b/mg)1/2v} ].
The second form we can convert from t(v) into v(t):
v(t) = (mg/b)1/2 tanh[(gb/m)1/2 t] = vt tanh[(gb/m)1/2 t] ] = vt tanh[gt/vt] since (mg/b)1/2 = vt.
To get x(t), we have v(t) = dx/dt = vt tanh[gt/vt] .
If we let u = gt/vt , we get:
[From the web,
(1/vt)*(x - xo) = (vt/g) o gt/v tanh(u) du
 tanh(u) du = ln(cosh[u]) + C ]
where cosh(u) = (eu + e-u)/2
x(t) = xo + (vt2/g) ln(cosh[gt/vt]) = xo + (m/b) ln(cosh[{bg/m}1/2t])
Limiting cases:
For t=0, cosh(0) = 1, and ln(1) = 0, so x(t=0) = xo as it should.
For small x, cosh(x) ≈ 1 + x2/2, and log(1+x) ≈ x, so
x(small t) ≈ xo + (m/b) (bg/m) t2/2 = xo + ½ g t2
(since the object falls with little air resistance).
For large x, cosh(x) ≈ ½ ex and ln(½ ex) = x – ln(2) which approaches just x, so
x(large t) ≈ xo + (m/b) * (bg/m)1/2t = xo + (mg/b)1/2t = xo + vt t
(since v approaches the constant terminal speed) .
Example (with units): For a baseball, with m = .145 kg, b = 7.6x10-4 kg/m, and
g = 9.8 m/s2 in a time t = 4 seconds, the ball will have travelled: x(4 sec) =
(.145 kg / 7.6x10-4 kg/m) * ln(cosh[{(7.6x10-4 kg/m)*(9.8 m/s2)/(.145 kg)}1/2*(4 s)])
= (190.789 m)*ln(cosh[.907]) = (190.789 m)*ln(1.44) = 69.57 m = x(4 sec).
If we ignored air resistance, then to fall 69.57 meters would take
x = ½ g t2 so t = {2*x/g}1/2 . so t = {2*(69.57m)/(9.8 m/s2)}1/2 = 3.77 seconds.
The time with air resistance is more than the time without as we would expect.
so we get:
It’s speed after falling 4 seconds (from a height of 69.57 m) would be:
v(t) = (mg/b)1/2 tanh[(gb/m)1/2 t]
v(4 sec) =
{(.145 kg)*(9.8 m/s2)(7.6x10-4 kg/m)}1/2 * tanh[{(7.6x10-4 kg/m)*(9.8 m/s2)/(.145 kg)}1/2*(4 s)])
= (43.24 m/s) * tanh(.907) = (43.24 m/s)*(.719) = 31.1 m/s.
Note that this speed is 71.9% of its terminal speed.
If we ignored air resistance, after falling 3.77 seconds from a height of 69.57 m, the ball would
have been going v = g*t = (9.8 m/s2)*(3.77 s) = 36.9 m/s.
The speed with air resistance is less than the speed without as we would expect.
Remember: in this case we are treating down as the positive direction, so positive x’s are lower
than where we started, and positive speeds are directed down.
Case 3: Going UP with FAR = - bv2 .
F = ma leads to -mg – bv2 = m dv/dt ,
Separating the variables gives:
Here we define up as positive.
(- dt) = (m dv) / (mg+bv2) = (1/g)(dv) / [1 + (b/mg)v2]
Now let u = (b/mg)1/2 v so dv = (mg/b)1/2 du :
(-g)*t = (mg/b)1/2 uo u (1 / [1 + u2]) du
The integral can be found using the following trig substitution:
Let u = tan (θ) so that du/dθ = 1 + tan2(θ) so du = [1 + tan2(θ) ] dθ
(-g)*t = (mg/b)1/2 θo θ dθ
which gives -(bg/m)1/2 t = θ – θo .
With θ = tan-1(u) and u = (b/mg)1/2v, this leads to:
t(v) = (m/bg)1/2 [ tan-1({b/mg}1/2 vo) - tan-1({b/mg}1/2 v) ] .
This can be converted from t(v) to v(t):
v(t) = (mg/b)1/2 tan[ tan-1({b/mg}1/2vo) – {bg/m]1/2t ]
Limiting cases:
For t=0, v(0) = vo as required.
For v=0 (highest point, beyond which this case no longer applies):
Highest Point: t(v=0) = thighest = (m/bg)1/2 [ tan-1({b/mg}1/2 vo)] .
For small θ (vo small) , tan(θ) ≈ θ ≈ tan-1(θ) so thighest ≈ (m/bg)1/2 ({b/mg}1/2 vo) = vo/g
which is what we would expect if there is no air resistance (v = vo – gt, so thighest = vo/g ).
Example (with units): If you throw a baseball up at 100 mph (44.64 m/s), how long will it take
to reach the highest point? For a baseball, with m = .145 kg, b = 7.6x10-4 kg/m, and g = 9.8
m/s2 : thighest = (m/bg)1/2 * tan-1({b/mg}1/2 vo) =
{(.145 kg)/[(7.6x10-4 kg/m)*(9.8 m/s2)]}1/2 * tan-1[ {(7.6x10-4 kg/m)/[(.145 kg)*(9.8 m/s2)]}1/2 * (44.64 m/s)]
= {4.412 sec} * tan-1[(.023 s/m)*(44.64 m/s)] = {4.412 sec} * tan-1[1.032]
= {4.412 sec} * [45.91o * (pi/180o)] = 3.54 seconds.
For comparison, with no air resistance, v = vo – g*t, so with v=0,
thighest = vo/g = (44.64 m/s) / (9.8 m/s2) = 4.56 seconds.
The time with air resistance is less than the time without as we would expect.
Case 3: Going up with FAR = - bv2 (extended)
Now that we have v(t), can we find x(t) for the case of being projected upwards with air
resistance and near-earth gravity?
v(t) = (mg/b)1/2 tan[ tan-1({b/mg}1/2vo) – {bg/m]1/2t ] = dx(t)/dt
To simplify the intermediate steps, lets define C = tan-1({b/mg}1/2vo) and separate variables:
(b/mg)1/2 dx = tan{ C – (bg/m)1/2t} dt .
Let’s now make the substitution: u = C – (bg/m)1/2t so that du/dt = -(bg/m)1/2 and so
dt = -(m/bg)1/2 du. We now have
(b/mg)1/2 dx = -(m/bg)1/2 tan(u) du, or
-(b/m) xo x dx =
uo
u
tan(u) du .
 tan(u) du =  [sin(u)/cos(u)] du ;
if we let w = cos(u), then dw/du = -sin(u), so sin(u) du = -dw, and so we get
 [sin(u)/cos(u)] du =  [-1/w] dw = - ln(w) (with both upper or lower limits inserted)
Thus, we get:
-(b/m)(x-xo) = - ln(cos[C-{bg/m}1/2t]) + ln(cos[C]) . This becomes:
x(t) = xo + (m/b) * ln { (cos[tan-1({b/mg}1/2vo – {bg/m}1/2t] ) / (cos[tan-1({b/mg}1/2vo)] ) } .
Limiting Cases:
For t=0, x = xo as required.
This solution only applies for 0 < t < thighest , so there is no looking at what happens as t goes to
infinity.
Example with units: For a baseball, with m = .145 kg, b = 7.6x10-4 kg/m, and
g = 9.8 m/s2 that is thrown upwards with a speed of 100 mph (44.64 m/s), how high will the ball
go? We saw in the above Case 2 that the time to reach the highest point is 3.536 seconds.
First let’s evaluate
(1) (m/b) = (.145 kg) / (7.6x10-4 kg/m) = 190.789 m;
(2) tan-1({b/mg}1/2vo) = tan-1({(7.6x10-4kg/m)/[(.145 kg)*(9.8 m/s2)]}1/2*(44.64 m/s))
= tan-1 (1.032) = 45.91o ;
(3) {bg/m}1/2*t = {(7.6x10-4 kg/m)*(9.8 m/s2) / (.145 kg)}1/2 * 3.536 s = .801 radian * (180o/pi
radians) = 45.91o This shouldn’t be a surprise, since we determined thighest based on {bg/m}1/2*t
= tan-1({b/mg}1/2vo) earlier.
xhighest = 0m + 190.789 m * ln {cos(0o) / cos(45.91o) } = 190.789 m * ln {1.439} = 69.4 m.
A numerical calculation with dt=0.02 seconds (see spreadsheet via link) gives a maximum height
of 69.0 meters at a time of 3.50 seconds.
As a comparison, without air resistance, the ball would go a height h:
½*m*vo2 = m*g*h, or h = vo2/(2*g) = (44.64 m/s)2 / (2 * 9.8 m/s2) = 101.7 m.
With air resistance, the ball does not go as high as it would without air resistance, as expected.
Note: This height is almost the same height as in the example for case 2, where we started from
a height of 69.6 m. It took 3.54 second to go up to that height with an initial speed of 44.64 m/s,
and it took a longer 4.0 seconds to go back down and it ended up with a lower speed of 31.1 m/s.
The above example does NOT include effects of a spinning ball nor of any updrafts or
downdrafts in the air.