SME
University Physics I: Mechanics · Problem type
Falling against linear drag: terminal speed
Use this for “A 2 kg ball is dropped from rest, and the air pushes back with 4 kg/s times its speed. Find its terminal speed and how fast it falls after 1 s.”
Chart
v(t) = 4.91(1 − exp(−t/τ)), τ = 0.5 s
After τ the response has made 63.2% of its change; after 5τ, 99.3%.
At t = 1 s: v = 4.241 m/s.
This is an example. Type your own values to start a new problem.
vT (m/s)
=
m (kg)
×
9.81
b (kg/s)
τ
Time constant
calculated
t
Time after release
example
v
Speed at t
calculated
Formulas
vT
=
m
×
9.81
b
4.905
=
2
×
9.81
4
τ
=
m
b
0.5
=
2
4
v
=
vT
×
(1
−
e
−
t
÷
τ
)
4.2412
=
4.905
×
(1
−
e
−1 ÷ 0.5
)
Assumptions
•
Drag is bv, in proportion to the speed (slow, small objects); g = 9.81 m/s².
•
The object starts from rest at t = 0 and falls straight down.
•
The terminal speed vT is where drag equals weight: bvT = mg, so it stops speeding up.
•
τ = m ÷ b; after 5τ the speed is within 1% of vT.
Step-by-step
WE KNOW
m = 2 kg, b = 4 kg/s, t = 1 s
FIND
terminal speed (vT), time constant (τ), speed at t (v)
Step 1 · Find terminal speed (vT)
Use vT = m × 9.81 ÷ b
At
terminal
speed
drag
balances
weight,
bvT
=
mg,
so
divide
the
weight
by
b
.
vT
=
2
×
9.81
4
vT
=
19.62
4
vT = 4.905 m/s
Step 2 · Find time constant (τ)
Use τ = m ÷ b
Newton’s
second
law,
m
dv/dt
=
mg
−
bv,
changes
the
speed
on
the
time
scale
m
b
.
τ
=
2
4
τ
=
0.5
s
Step 3 · Find speed at t (v)
Use v = vT × (1 − e−t ÷ τ)
Solving
m
dv/dt
=
mg
−
bv
from
rest
gives
v
=
vT(1
−
e
−
t
τ
):
the
share
of
vT
reached
so
far.
v
=
4.905
×
(1
−
e
−1 ÷ 0.5
)
v
=
4.905
×
(1
−
e
-2
)
v
=
4.905
×
(1
−
0.1353)
v
=
4.905
×
0.8647
v
=
4.2412
m/s
Check
Put the numbers back in. Both sides should match.
4.905
=
2
×
9.81
4
✓
0.5
=
2
4
✓
4.2412
=
4.905
×
(1
−
e
−1 ÷ 0.5
)
✓
← Previous lesson
An Atwood machine: two hanging blocks
Next lesson →
A banked curve with no friction: the design speed
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
Related lessons
Newton's laws
Main lesson
An Atwood machine: two hanging blocks
Use this for “Blocks of 3 kg and 5 kg hang from the two ends of a string over a pulley. Find their acceleration and the string’s tension.”
A banked curve with no friction: the design speed
Use this for “A curve of radius 80 m is banked at 12°. At what speed can a 1500 kg car take it with no friction, and how hard does the road push on it?”