Hello Music Freak,
I believe there to be a typo in your statement of "equation 4."
I would begin with Newton's second law of motion:
$$ma=F$$
Since acceleration $a$ is defined to be the time rate of change of velocity $v$, we may write:
$$m\frac{dv}{dt}=F$$
We have two forces acting on the paratrooper...the force of gravity and drag, which we are told is proportional to his velocity. The force of gravity is in the same direction as his velocity (both are in the downward direction), which the force of drag, opposing the motion, points upwards, hence:
$$m\frac{dv}{dt}=mg-kv$$
where $0<k$ is called the coefficient of drag. Now, if we divide through by the mass of the paratrooper, we obtain:
$$\frac{dv}{dt}=g-\frac{k}{m}v$$
If we define $$\rho=\frac{k}{m}$$ then we may write:
$$\frac{dv}{dt}=g-\rho v$$ where $$v(0)=v_0$$
Writing the ODE in linear form, we have:
$$\frac{dv}{dt}+\rho v=g$$
Computing the integrating factor, there results:
$$\mu(t)=e^{\rho\int\,dt}=e^{\rho t}$$
and so the ODE becomes, after multiplying through by this factor:
$$e^{\rho t}\frac{dv}{dt}+\rho e^{\rho t}v=ge^{\rho t}$$
Now, the left side may be expressed as the derivative of a product:
$$\frac{d}{dt}\left(e^{\rho t}v \right)=ge^{\rho t}$$
Integrating with respect to $t$, we have:
$$\int\,d\left(e^{\rho t}v \right)=g\int e^{\rho t}\,dt$$
$$e^{\rho t}v=\frac{g}{\rho}e^{\rho t}+C$$
Dividing through by $e^{\rho t}$, we obtain the general solution:
$$v(t)=\frac{g}{\rho}+Ce^{-\rho t}$$
Using the initial condition, we may determine the value of the parameter $C$:
$$v(0)=\frac{g}{\rho}+C=v_0\,\therefore\,C=v_0-\frac{g}{\rho}$$
and so the solution to the IVP is:
$$v(t)=\frac{g}{\rho}+\left(v_0-\frac{g}{\rho} \right)e^{-\rho t}$$
Now, it is easy to see that the terminal velocity is:
$$v_{\max}=\lim_{t\to\infty}v(t)=\frac{g}{\rho}$$
Thus, we may state the solution as:
$$v(t)=v_{\max}+\left(v_0-v_{\max} \right)e^{-\frac{g}{v_{\max}} t}$$
Next, using the fact that the time rate of change of position (or distance fallen at time $t$ which we'll call $x(t)$) is velocity, we have the IVP:
$$\frac{dx}{dt}=v(t)=v_{\max}+\left(v_0-v_{\max} \right)e^{-\frac{g}{v_{\max}} t}$$ where $$x(0)=x_0$$
Treating $dx$ and $dt$ as differentials, we may separate variables to obtain:
$$dx=\left(v_{\max}+\left(v_0-v_{\max} \right)e^{-\frac{g}{v_{\max}} t} \right)\,dt$$
$$\int\,dx=\int v_{\max}+\left(v_0-v_{\max} \right)e^{-\frac{g}{v_{\max}} t}\,dt$$
$$x(t)=v_{\max}t+\frac{v_{\max}\left(v_{\max}-v_0 \right)}{g}e^{-\frac{g}{v_{\max}} t}+C$$
Using the initial condition, we may determine the value of the parameter $C$:
$$x(0)=\frac{v_{\max}\left(v_{\max}-v_0 \right)}{g}+C=x_0\,\therefore\,C=x_0-\frac{v_{\max}\left(v_{\max}-v_0 \right)}{g}$$
and so the solution to the IVP is:
$$x(t)=v_{\max}t+\frac{v_{\max}\left(v_{\max}-v_0 \right)}{g}e^{-\frac{g}{v_{\max}} t}+x_0-\frac{v_{\max}\left(v_{\max}-v_0 \right)}{g}$$
$$x(t)=v_{\max}t+\frac{v_{\max}\left(v_{\max}-v_0 \right)}{g}\left(e^{-\frac{g}{v_{\max}} t}-1 \right)+x_0$$
Now, if we use:
$$v_0=0\frac{\text{ft}}{\text{s}}$$
$$x_0=0\text{ ft}$$
$$v_{\max}=100\frac{\text{mi}}{\text{hr}}\cdot\frac{5280\text{ft}}{1\text{ mi}}\cdot\frac{1\text{ hr}}{3600\text{ s}}=\frac{440}{3}\,\frac{\text{ft}}{\text{s}}$$
$$\frac{g}{v_{\max}}=\frac{32.17405\dfrac{\text{ft}}{\text{s}^2}}{\dfrac{440}{3}\, \dfrac{\text{ft}}{\text{s}}}=\frac{1930443}{8800000}\,\frac{1}{\text{s}}$$
$$t=8\text{ s}$$
We then have:
$$x\left(8\text{ s} \right)=\left(\frac{440}{3}\,\frac{\text{ft}}{ \text{s}} \right)\left(8\text{ s} \right)+\frac{\left(\dfrac{440}{3}\,\dfrac{\text{ft}}{\text{s}} \right)^2}{\dfrac{643481}{20000}\,\dfrac{\text{ft}}{\text{s}^2}}\left(e^{-\left(\dfrac{1930443}{8800000}\,\dfrac{1}{\text{s}} \right)\left(8\text{ s} \right)}-1 \right)$$
$$x\left(8\text{ s} \right)=\left(\frac{3520}{3}+\frac{38720000000}{57911329}\left(e^{-\dfrac{1930443}{1100000}}-1 \right) \right)\text{ ft}\approx620.338173295040\text{ ft}$$
So, after 8 seconds, the paratrooper would have fallen only about 620 ft. To find the time required to fall 1200 ft, we may set:
$$x(t)=1200\text{ft}$$
$$1200\text{ ft}=\left(\frac{440}{3}\,\frac{\text{ft}}{ \text{s}} \right)\left(t\text{ s} \right)+\frac{\left(\dfrac{440}{3}\,\dfrac{\text{ft}}{\text{s}} \right)^2}{\dfrac{643481}{20000}\,\dfrac{\text{ft}}{\text{s}^2}}\left(e^{-\left(\dfrac{1930443}{8800000}\,\dfrac{1}{\text{s}} \right)\left(t\text{ s} \right)}-1 \right)$$
Using a numeric root finding technique, since we cannot solve for $t$ explicitly, we find:
$$t\approx12.442929881626764089\text{ s}$$
