Writing Newton's second law
If you dropped a little rock from the top of a small mountain at not too high of an altitude, what would happen to that rock during its descent? To answer this question, we need to think about what forces are acting on the rock. Afterall, the net force acting on an object is what determines what that object's motion will look like. There are two forces acting on the rock. First, there is the force of gravity \(\vec{F}_g\); since the altitude of a small mountain (or indeed any mountain) is much, much smaller than the Earth's radius, we can assume that \(Δy<<R_E\) where \(Δy\) is the change in altitude of the rock and \(R_E\) is the radius of the Earth. Thus, \(F_g=+mg\) if we choose the direction of the positive x-axis to point down towards the ground. As the rock falls through the Earth's atmosphere, all of the Earth's air molecules are constantly colliding with the rock. These molecular collisions against the rock are the microscopic origin of the second force acting on the rock: the force of air friction. The force of gravity is very simple as long as \(Δy<<R_E\): it's just the constant \(mg\). The force of air friction can, in general, be very complicated. But if an object is moving at a slow enough velocity and if its change in altitude isn't very large, the force of air friction can be approximated as
$$F_d=-bv,\tag{1}$$
where \(b\) is a constant called the drag coefficient.
To describe the rock's motion we'll derive expressions for the functions \(a(t)\) (if we were ignoring air friction, then \(a(t)\) would simply just be the constant \(g\)), \(v(t)\), and \(y(t)\) by using Newton's second law. What will the expression for the net force, \(\sum{F}\), acting on the rock look like? Since we've assumed that \(Δy<<R_E\), we know that the magnitude and direction of the force \(\vec{F}_g\) will stay constant. What is the sign of this forces component, \(F_g\)? The choise of sign is completely arbitrary but since we typically, by convention, choose the "positive direction" to be the same direction as the object's motion, we'll choose the sign of \(F_g\) to be positive. Thus, \(F_g=+mg\). (As a side note, we could choose the sign to be negative and derrive all the same results for \(a(t)\), \(v(t)\), and \(y(t)\).)
The force of friction always acts in a direction opposite to that of an object's motion. Thus, the sign of \(F_d\) must be negative and \(F_d=-bv\). Since these are the only two forces acting on the rock, the net force must be
$$\sum{F}=mg-bv=ma.\tag{2}$$
Finding the terminal velocity
As the rock is falling, the force of friction exerted on the object will keep increasing. But eventually, after a certain amount of time \(t_T\) has passed, the force \(-F_d\) will become perfectly counter-balanced by the force \(F_g\) and the two forces will cancel. Thus, at the time \(t_T\), the net force acting on the object is zero. Consequently, by Newton's second law, \(a(t)=0\) such that \(tgreater than or equal to t_T\). With no more acelleraation, the velocity of the object becomes constant. This velocity is know as the terminal velocity which we'll represent by \(v_T\). This simplifies Equation (2) to
$$mg-bv_T=0.$$
Using algebra, we can determine the object's terminal velocity:
$$v_T=\frac{mg}{b}.\tag{3}$$
Finding the equations of motion
Let's rewrite Equation (2) as
$$mg-bv=m\frac{dv}{dt}.\tag{4}$$
Equation (4) is a first-order seperable differential equation. To solve such a differential equation we must get all of the "\(v\) terms" on one side of the equation and all of the "\(t\) terms" on the other side. To do that, we must do the following algebra:
$$mg-bv=m\frac{dv}{dt}$$
$$(mg-bv)dt=mdv$$
$$dt=\frac{m}{mg-bv}dv.\tag{5}$$
The next step is to integrate both sides of Equation (5) to get
$$\int_0^tdt=m\int_0^v\frac{1}{mg-bv}dv.$$
How do we solve the integral above? Well, if I asked you to solve the integral,
$$\int_{\text{some amount}}^{\text{some other amount}}\frac{1}{\text{"some variable}}d(\text{"some variable")},$$
you would tell me that the solution to the integral is just \([ln|\text{"some variable"}|]_{\text{"some amount"}}^{\text{"some other amount"}}\). If we define some new variable \(u\) as \(u=mg-bv\), we can essentially make that simplification and solve the integral. Letting \(u=mg-bv\), we can do algebra to find that
$$\frac{du}{dv}=\frac{d}{dv}(mg-bv)$$
$$\frac{du}{dv}=-b$$
$$du=-bdv$$
$$dv=\frac{-1}{b}du.\tag{6}$$
Substituting \(u=mg-bv\) and Equation (6) into the integral, the integral simplifies to
$$t=-\frac{m}{b}\int_{?_1}^{?_2}\frac{1}{u}du.\tag{7}$$
What should the limits of integration be for Equation (7)? To find the limits of integration, we can use the equation \(u=mg-bv\). The lower limit, \(?_1\), is the value of \(u\) when \(v=0\) which is given by \(?_1=u=mg\). The upper limit, \(?_2\), is the value of \(u\) when \(v=v\) which is given by \(?_2=u=mg-bv\). Substituting these values into Equation (7), we have
$$t=-\frac{m}{b}\int_{mg}^{mg-bv}\frac{1}{u}du.\tag{8}$$
Solving the integral in Equation (8) and doing some algebra, we have
$$t=-\frac{m}{b}[ln|u|]_{mg}^{mg-bv}$$
$$-\frac{bt}{m}=ln|mg-bv|-ln|mg|=ln|\frac{mg-bv}{bg}|$$
$$e^{-bt/m}=\frac{mg-bv}{mg}$$
$$mge^{-bt/m}=mg-bv$$
$$v=mg-mge^{-bt/m}=\frac{mg}{v}(1-e^{-bt/m})$$
$$v(t)=v_t(1-e^{-bt/m})\tag{9}$$
To solve for the motion \(y(t)\), let's integrate both sides of the equation above with respte to time to get
$$y(t)=\int_0^tv_T(1-e^{-bt/m})dt=v_T\int_0^tdt-\int_0^te^{-bt/m}dt$$
To solve the integral on the right-hand side, let \(u=-bt/m\). Thus,
$$\frac{du}{dt}=-\frac{b}{m}$$
$$du=-\frac{b}{m}dt$$
$$dt=-\frac{m}{b}du.$$
Making the appropriate substitutions, we can simplify the expression for \(y(t)\) to
$$y(t)=v_Tt+\frac{b}{m}\int_0^{-bt/m}e^udu=v_Tt+\frac{b}{m}e^u|_0^{-bt/m}=v_Tt+\frac{b}{m}(e^{-bt/m}-1)$$
We can also take the derrivative of both sides of Equation (9) to find \(a(t):
$$a(t)=\frac{d}{dt}(v_T-v_Te^{-bt/m})=-v_Te^{-bt/m}(\frac{-b}{m})=\frac{mg}{b}\frac{b}{m}e^{-bt/m}=ge^{-bt/m}$$