For the case B=0
we are left with the differential equation given by
dv/dt = -g + Av
which has the well-known solution for an object which moves
in the vertical direction under the influence of the gravitational
attraction near the surface of the Earth
and, moreover, under the influence of a frictional force
proportional to its velocity
v = v(t) = vterm(
e- t/T - 1 )
Here vterm>0
and T>0 represent constants in time
(i.e. positive real numbers independent of the instant of time
t) which will be interpreted below.
Notice that T has units
seconds
and vterm
units m/s.
In the figure below we show how the expression
e- t/T
varies with time t.
e- t/T
as function of t
for T=1/2 s,
T=1 s
and T=2 s.
We observe that
e- t/T>0
for any instant of time t
and, furthermore, that
e- t/T>1
for t<0,
whereas
e- t/T<1
for t>0.
As a consequence
e- t/T-1>0
for t<0,
hence v>0
(upward motion),
whereas
e- t/T-1<0
for t>0,
hence v<0
(downward motion).
At t=0
we obtain v=0.
Hence, our solution describes an object which reaches its maximum height
at t=0.
Furthermore we observe that
e- t/T≅0
for t>>T.
Hence
v(t>>T) ≅ -vterm
The object moves with constant downward velocity,
because the gravitational acelleration -g
is compensated by the frictional acelleration
Avterm.
by substitution of the solution for v
in the above differential equation.
Notice that when A
is a negative constant
the acelleration Av
is always opposite
to the direction of the velocity,
right as it should be.
