MATH 137
The Falling Object Problem
With Air Resistance
The Problem: Suppose we drop an object from initial height h0 with no initial velocity.
The object is acted upon only by gravity that is dampened by air resistance.
(i)
(ii)
(iii)
(iv)
What is the object’s velocity in t seconds?
What is the object’s terminal speed?
How long will it take to hit the ground?
What are the object’s general height and velocity functions?
Throughout, we shall let g be the acceleration due to gravity on Earth, which we
may take to be g ≈ 9.8 m/s2 or g ≈ 32.174 ft/s2.
The height is measured positively from the ground upward. So a falling object has
a decreasing height, which makes its velocity negative. Because the velocity is becoming
more negative as the object falls, the acceleration also is negative. Thus, we must use
the value – g to denote that gravity is pulling the object downward. However, for a
falling object, the air resistance acts in the opposite direction as gravity. That is, the air is
trying to prevent the fall of the object.
The Air Resistance Equations
A body falling through air causes turbulence that results in air resistance that is
proportional to the square of the velocity. For an object going downward, the air resistance
is in the opposite direction of gravity; hence, the acceleration downward is
2
a(t) = − g + k(v(t ))
m 
ft 
2  or 2  , for 0 ≤ t ≤ T f ,
s
s
(1)
where T f in seconds is the time it takes to hit the ground.
€
The constant k depends on the shape, orientation, and mass of the object, as well as
the density of the atmosphere. For example, a baseball around sea level has a constant
€ value of about k ≈ 0.00187 per foot, or k ≈ 0.006125 per meter.
By separating variables in Equation (1) and integrating, we obtain the following
velocity function:
v(t ) = −
m 
ft 
g
 or  , for 0 ≤ t ≤ T f .
tanh ( (kg) t )
s 
s
k
(2)
By taking the limit of v(t ) as t → ∞, the terminal velocity of a falling object is found
€
to be – g / k m / s (or ft s ). If we can measure the terminal speed sT of an object, then
we find its constant k by solving sT = g / k to obtain
k =
€
g
(sT ) 2
per meter (or per foot).
(3)
The following chart gives some approximate terminal speeds and constants k .
Body
Terminal Speed
Skydiver (Vertical)
Skydiver (spread eagle)
Parachutist
Ping Pong Ball
Baseball
Golfball
Stone (1 cm rad)
Raindrop
85 m/s
55 m/s
6.5 m/s
7.0 m/s
40 m/s
30 m/s
30 m/s
10 m/s
k
278.8 ft/s
180.4 ft/s
21.32 ft/s
22.96 ft/s
131.2 ft/s
98.4 ft/s
98.4 ft/s
32.8 ft/s
0.0013564/m
0.000414 /ft
0.00324 /m
0.000987 /ft
0.23195 /m
0.0708 /ft
0.2 /m
0.06103 /ft
0.006125 /m
0.00187 /ft
0.01089 /m
0.003323 /ft
0.01089 /m
0.003323 /ft
0. 098/m
0.0299 /ft
Now by integrating the velocity function in (2), we obtain the height function:
1
h(t) = h0 − ln cosh (kg) t
k
(
(
)) meters (or ft) ,
1
where h0 is the initial height. So ln cosh (kg) t
k
€
( (
for 0 ≤ t ≤ T f ,
)) gives the actual distance fallen.
€
(k g) = g 2 / (sT ) 2 = g /sT . Equations (2) and (4) can
Note: Because k = g / (sT ) 2 , then
then be written as
€
(s ) 2
m 
ft 
€
 or  and h(t) = h0 − T ln(cosh((g /sT ) t ))
v(t) = −sT tanh((g /sT€) t)
g
s 
s
for 0 ≤ t ≤ T f .
€
€
Example 1. A baseball is dropped from €
20 meters.
(i) What are the baseball’s general height and velocity functions?
(ii) How long will it take to hit the ground, and what is the impact speed?
Solution. (i) Using k ≈ 0.006125 per meter and g ≈ 9.8 m/s2 , we have
ln(cosh(0.245 t ))
1
h(t) = h0 − ln cosh (kg) t ≈ 20 −
meters,
k
0.006125
(
))
(
and
€
v(t) = −
(4)
g
tanh (kg) t = −40 tanh(0.245 t ) m/s.
k
(
)
(ii) In general, the time needed to hit the ground is found by solving h(t) = 0 :
€
(5)
h0 −
1
1
ln ( cosh ( (kg) t )) = 0 →
ln ( cosh ( (kg) t ) ) = h0 → ln ( cosh ( (kg) t )) = k h0
k
k
(kg) t = cosh−1 (e k h0 ) → t =
→ cosh ( (kg) t ) = e k h0 →
In this case, the required
€ time is t ≈
cosh−1 (e k h0 )
sec
(kg)
cosh−1 (e 0.006125× 20 )
≈ 2.0618 sec.
(0.006125€
× 9.8)
At this time, the impact velocity is −40 tanh(0.245 × 2.0618) ≈ −18.646 m/s .
€
€ an object has a terminal speed of about 96 mph. (i) What
Example 2. We measure that
is the object’s air resistance constant k ? (ii) Two seconds after dropping it, what is its
speed and how far does it fall?
Solution. (i) We first convert the speed to ft/sec:
sT = 96
miles
ft
1 hr
ft
× 5280
×
= 140.8
hr
mile 3600 sec
sec
Then
€
k=
g
=
(sT ) 2
32.174 ft /sec 2
140.8 2 ft 2 /sec 2
≈ 0.001623 per foot .
(Multiplying by 3.28 ft/meter, we obtain k ≈ 0.0053232 per meter.)
€
€
(ii) Now v(t) = −sT tanh((g /sT ) t) ≈ −140.8 tanh((32.174 /140.8) t) ft/s, so that in 2 seconds
€
after falling the velocity is
€
v(2) = −140.8 tanh((32.174 /140.8)× 2) ≈ −60.2131 ft/s, or about 41 mph downward.
€
The actual distance that an object falls in t seconds is given by
€
1
ln cosh (kg) t
k
( (
))
=
(sT ) 2
ln(cosh((g /sT ) t )) ft (or meters)
g
€
In this case, the object falls
€
140.8 2
ln(cosh( 32.174 /140.8 × 2)) ≈ 62.225 ft
32.174
€
€
Example 3. An object takes 0.872 seconds to fall 10 feet after dropping it. Solve for the
object’s air resistance constant k . Then give the terminal speed of the object.
Solution. The falling distance in t seconds is
for k in the equation
1
ln cosh (32.174 k) t . So we must solve
k
( (
))
1
ln cosh (32.174 k) × .872 = 10 . That is, we must solve
k
€ 1
ln cosh (32.174
k) × 0.872 −10 = 0 .
€
k
( (
( (
))
))
€ there is no algebraic solution for k . Thus we must solve for k using
Unfortunately,
numerical methods.
€
We find that k ≈ 0.0621 per foot. The terminal speed of the object is then
sT =
g/ k ≈
32.174
≈ 22.76 ft/sec
0.0621
Exercises
1. An object takes 2.7 seconds to fall 60 feet to the ground when dropping it with no
initial velocity.
(a) Find its air resistance constant k .
(b) Find its terminal speed in mph.
(c) Find its speed in mph when it hits the ground.
2. A golfball is dropped from 50 feet.
(a) What are it’s general height and velocity functions?
(b) How long will it take to hit the ground, and what is the impact speed?
3. When a baseball is dropped from up high, how long does it take for it to reach 99%
of its terminal velocity?
4. An object falls 8 meters in 1.44 seconds when dropping it with no initial velocity.
(a) Find its air resistance constant k and its terminal speed.
(b) How far would it fall in 2 seconds after dropping it?
(c) Determine its velocity and acceleration at 2 seconds after dropping it.
Solutions
1. (a) We must solve for k in the equation
must solve
1
ln cosh (32.174 k) × 2.7 = 60 . That is, we
k
( (
))
1
ln cosh (32.174 X) × 2.7 − 60 = 0 .
X
€
( (
))
€
We find that k ≈ 0.0387 per foot. (b) The terminal speed of the object is then
sT =
g/ k ≈
32.174
ft
≈ 20.83
× 3600 / 5280 ≈ 19.66 mph
0.0387
sec
(c) The impact speed at 2.7 seconds is
€
32.174
tanh (.0387 × 32.174) × 2.7 = 28.69 ft/s or 19.56 mph
.0387
----------------------------------------------------------------
(
)
2. (a) Using k ≈ 0.003323 per ft and g ≈ 32.174 ft/s2 , we have
€
ln(cosh(0.327 t ))
1
h(t) = h0 − ln cosh (kg) t ≈ 50 −
ft,
k
0.003323
and
g
v(t) = −
tanh (kg) t = −98.4 tanh(0.327 t ) ft/s.
k
(
(
€
))
(
)
−1 0.003323 ×50
)
(b) Solving h(t) = 0 gives t = cosh (e
≈ 1.812 sec.
0.327
€
€
Then v(1.812) ≈ −98.4 tanh(0.327 ×1.812) ≈ −52.32 ft/sec (≈ 35.673 mph)
- - -€- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - €
3. For any object, solve v(t) = −sT tanh((g /sT ) t) = – 0.99 sT . Thus,
g
s
t = tanh−1 (0.99) . Hence, t = T tanh−1 (0.99) .
tanh((g /sT ) t) = 0.99 or
sT
g
€
€
For a baseball, sT ≈ 40 m /s , so the time required is
€
€
€
€
s
40€ −1
t = T tanh−1 (0.99) =
tanh (0.99) ≈ 10.8 sec .
g
9.8
4. (a) We must solve for k in the equation
must solve
1
ln cosh (9.8 k) ×1.44 = 8 met . That is, we
k
( (
))
1
ln cosh (9.8 X) × 1.44 − 8 = 0 .
X
€
( (
))
€
We find that k ≈ 0.09272 per meter. The terminal speed of the object is then
sT =
9.8
≈ 10.28 met/sec.
0.09272
g/ k ≈
1
(b) The distance fallen in two seconds is
ln cosh .09272 × 9.8 × 2 ≈ 13.32 met.
.09272
€
( (
))
(c) The velocity at 2 seconds is
€
9.8
v(2) = −
tanh (.09272 × 9.8) × 2 = −9.837 m/s .
.09272
(
)
The acceleration at time t is a(t) = −g + k(v(t)) 2 . So at 2 seconds the acceleration is
€
a(2) ≈ −9.8 + .09272 × (9.837) 2 = −0.8278 m/s2 .
€ €
There is almost no accelaration downward because the object has nearly reached
terminal velocity.
€