fisica1-youn-e-freedman-exercicios-resolvidos

5.124: For convenience, take the positive direction to be down, so that for the baseball
released from rest, the acceleration and velocity will be positive, and the speed of the
baseball is the same as its positive component of velocity. Then the resisting force,
directed against the velocity, is upward and hence negative.
a)
2
b) Newton’s Second Law is then ma = mg − Dv . Initially, when v = 0,
the
acceleration is g, and the speed increases. As the speed increases, the resistive force
increases and hence the acceleration decreases. This continues as the speed approaches
mg
the terminal speed. c) At terminal velocity, a = 0,
so v = , in agreement with
dv g 2 2
Eq. (5.13). d) The equation of motion may be rewritten as = ( − v ). This is a
separable equation and may be expressed as
dv g
∫ =
2 2 2
v − v v
t
1 ⎛ v ⎞ gt
arctanh ,
2
v
⎜ =
t
v
⎟
⎝ t ⎠ v
t
so v = vt
tanh( gt vt
).
Note: If inverse hyperbolic functions are unknown or undesirable, the integral can be
done by partial fractions, in that
1 1 ⎡ 1 1 ⎤
=
,
2 2 ⎢ + ⎥
vt
− v 2vt
⎣vt
− v vt
+ v⎦
and the resulting logarithms in the integrals can be solved for v (t)
in terms of
exponentials.
t
∫
dt,
or
t
dt
D
2
vt
v t