# If the effects of atmospheric resistance are accounted for

If the effects of atmospheric resistance are accounted for, a falling body has an acceleration defined by the equation $a = 9.81[1-v^2(10^{-4})]$ m/$s^2$, where is v in m/s and the positive direction is downward. If the body is released from rest at a very high altitude, determine (a) the velocity when t = 5 s, and (b) the body’s terminal or maximum attainable velocity (as $t\rightarrow\infty$).

#### Solution:

Remember that we can write acceleration as:

$a(v)=\dfrac{dv}{dt}$

$dt=\dfrac{dv}{a(v)}$

Let us take the integral of both sides of the equation:

$\,\displaystyle \int^{t_2}_{t_1}dt=\int^{v}_{v_0}\dfrac{1}{a(v)}dv$

Substitute our acceleration equation:

$\,\displaystyle \int^{t}_{0}dt=\int^{v}_{0}\dfrac{1}{9.81[1-v^2(10^{-4})]}dv$

$t=\dfrac{50}{9.81}\text{ln}\left(\dfrac{1+0.01v}{1-0.01v}\right)$

Isolate for v:

$v=\dfrac{100(e^{0.1962t}-1)}{e^{0.1962t}+1}$

At t=5 seconds, we have:

$v=\dfrac{100(e^{(0.1962)(5)}-1)}{e^{(0.1962)(5)}+1}$

$v=45.5$ m/s

As $t\rightarrow\infty$, we have:

$v=100$ m/s

$v=45.5$ m/s
Maximum velocity, $v=100$ m/s