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Free fall with quadratic resistance
The following is a common cinematics problem in national and regional high school olympiads
...
The
gravity acceleration is g, and the air resistance is proportional to the square of the velocity, with
proportionality coefficient γ > 0
(https://en
...
org/wiki/Free_fall#Free_fall_in_Newtonian_mechanics)
...

X When will the point mass stop ascending?
X What is the maximal height reached?
X What is the asymptotics of the velocity?
X What is the velocity as a function of time?
X What is the position as a function of time?

We choose the coordinate axis x to point vertically upwards such that x(0) = 0
...
The air resistance force γv 2 is always directed
against the motion, hence the Newton’s second law is divided into two cases:

mv(t) =
˙



−mg − γv 2 (t)

if v(t) ≥ 0


−mg + γv 2 (t)

else

Thus the problem can be written as follows:
v(0) = v0 ,
v(t) = −g(1 +
˙

γ 2
v (t)),
mg

t ∈ (0, t0 ),

v(t0 ) = 0,
v(t) = −g(1 −
˙

γ 2
v (t)),
mg

1

t ∈ (t0 , +∞)
...


Introduce the following auxiliary function,
√
u(ξ) =

( √
)
γ
m
v −
ξ ,
mg
γg

√
v(t) =

ξ ∈ (−∞, 0],

( √
)
mg
γg
u −
t ,
γ
m

t ∈ [0, ∞)
...


To integrate the first equation write it as
dξ =

du
= d arctan u,
1 + u2

whence, using u(ξ0 ) = 0,
u(ξ) = tan(ξ − ξ0 ),

ξ ∈ [ξ0 , 0]
...

mg

Similarly, for the second equation,
dξ =

du
= d arctanh u,
1 − u2

thus (with u(ξ0 ) = 0)
u(ξ) = tanh(ξ − ξ0 ),

2

ξ ∈ (−∞, ξ0 ]
...
First, we find that
√
t0 =

m
arctan
γg

(√

)
γ
v0 ,
mg

the time when the point stops ascending
...


m

t ∈ [t0 , ∞),

and this gives us the velocity as a function of time
...
The asymptotic velocity is
√
lim v(t) = lim

t→∞

t→∞

(
(√
) √
)
√
mg
γ
γg
mg
tanh arctan
v0 −
t =−

...

0

Let us first find two indefinite integrals (α, β ∈ R, β ̸= 0),
∫
tan(α+βt)dt =
∫

1
β

∫
tan ξdξ = −

1
tanh(α + βt)dt =
β

∫

1
β

∫

d cos ξ
1
1
= − ln cos ξ = − ln cos(α+βt),
cos ξ
β
β

1
tanh ξdξ =
β

∫

d cosh ξ
1
1
= ln cosh ξ = ln cosh(α + βt)
...


In particular, the maximal achievable height is
m
h(t0 ) = h0 −
ln cos arctan
γ

(√

γ
v0
mg

)

(
)
2
m
γv0
= h0 +
ln 1 +
,
2γ
mg

where we used
1 + tan2 ξ =

1

...
We are done
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