Equations governing projectile motion
How do we describe the motion of a cannonball? More generally, how do we describe the motion of any object moving near the Earth's surface? The quest to answer these questions was a nearly 2,500 year long struggle which began with Aristotle. It wasn't until Newton and Liebnez invented calculus that we had the mathematics necessary to describe projectile motion. Whenever any object moves near Earth's surface, the entire planet Earth is exerting a "pull" and "tug" on that object in the downwards direction towards the ground. This "pull" always had the same strength and is always pointing down. The question as to why the Earth is constantly pulling objects down was first answered by Newton when he discovered the universal law of gravity. This law states that among all the trillions of galaxies that are visible to us (and perhaps, more, if there is more stuff out there), all of this matter, or "stuff," has a property called mass and is attracted to each other. For Newton, it must have had been an astonishing realization that a grain of dust in his room was pulling on another grain of dust on the other side off the universe—and, indeed, everything else in the universe.
We know from everyday experience that when we "pull" on something, that "pull" will, in general, effect that something's motion. We also know from common experience that there must be some sort of relationship between the strength of the pull and how much the objects motion changes. If I give a very small pull on a chair, it might move a little; but if I pulled on that same chair as hard as I possibly could, it might go flying across the room. We also know from everyday experience that how heavy or massive something is must also be related to its change in inertial motion. For example, if I pull on a small pebble, I can alter its inertial motion quite easily; but if I try pulling on the empire state building, it won't budge at all.
All of these quantities are relate to each other mathematically according to Newton's second law which states that the "pull" or "push," \(\vec{F}\), is equal to how massive the object is (which is represented by \(m\)) times its instantaneous acceleration: \(\vec{F}=m\vec{a}\). We can conclude from Newton's law of gravity and second law that: the Earth exerts a "pull" on the object, and that "pull" in the downward direction results in the object accelerating by an amount \(\vec{a}\) in the downward direction, which means that the object's velocity must be changing in the downward direction.
I tried to give somewhat of a friendly introduction to Newton's law of gravity and second law which, together, describes why the object falls down. These laws also tell us that the force \(\vec{F}_g\) that the Earth pulls down on objects with will always point down towards the ground. This means that the objects acceleration towards the ground (which we'll call \(\vec{g}\)) will also always point down and have the same constant magnitude.
It is good to have at least some knowledge that the proceeding results which we will derive using algebra an calculus are essentially coming from Newton's laws. Since, as I said in a previous lesson, kinematics is all about describing the motion of an object without consideration of what caused that motion (which was the force \(\vec{F}_g\)), we won't delve into any further details about why the object moves the way it does. We'll save that for the lessons on dynamics and mechanics.
The constant downward acceleration of any arbitrary object with any arbitrary trajectory near Earth's surface is given by
$$\frac{d\vec{v}}{dt}=\vec{g}=9.8\frac{m}{s}(-\hat{j}).\tag{1}$$
The acceleration \(\vec{g}(t)\) (it can be written as a function of time even though it's a constant) is a kind of "\(f'(x)\)-like function." As I discussed in my lessons on single-variable calculus, calculus is all about the math tools one has to use to go back and forth between \(f(x)\) and \(\frac{df}{dx}\) when \(\frac{df}{dx}\) is not a constant. We use integrals to go from \(\frac{df}{dx}\) to \(f(x)\) and derivatives to go from \(f(x)\) to \(\frac{df}{dx}\). Unfortunately, when \(f'(x)\) keeps changing, we have to use those two math tools to get the area or slope.
Our goal is to solve for the motion \(\vec{R}(t)\) of the object. If you notice this from the previous lesson, you'll get a hint as to how we can determine \(\vec{R}(t)\) given \(\vec{g}(t)\): notice that \(\vec{R}(t)\) and \(\vec{v}(t)\) are like \(f(x)\) and \(f'(x)\), and that \(\vec{v}(t)\) and\(\vec{g}(t)\) are like \(g(x)\) and \(g'(x)\).
In my lessons on single-variable calculus, I explain how to go from some function \(h'(x)\) to the function \(h(x)\), you have to find the area under \(h'(x)\). When \(h'(x)\) is a constant, finding the area is pretty simple. In Equation (1), the function \(\frac{d\vec{v}}{dt}\) is a constant (we'll explain why in detail when we cover dynamics) and the area under this function is just \(gΔt+C\). We can also take the integral on both sides of Equation (1) and use the rules of integral calculus (although there not necessary in this step to get
$$\int{\frac{d\vec{v}(t)}{dt}}dt=\int{\vec{g}(t)}dt$$
$$\vec{v}(t)=\vec{g}(t)t+c.\tag{2}$$
To solve for the constant \(C\), we can just set \(t=0\) giving us \(C=\vec{v}(0)\). Thus, \(C\) is equal to the initial velocity \(\vec{v}(0)\) which we'll represent by \(\vec{v}_0\). If we substitute for \(c\) into Equation (2) we'll have
$$\vec{v}(t)=\vec{v}_0+\vec{g}(t)t.\tag{3}$$
We have just one last step: we need to go from \(\vec{v}=\frac{d\vec{R}}{dt}\) to \(\vec{R}(t)\). Unlike when we went from \(\vec{a}=\frac{d\vec{v}}{dt}\) to \(\vec{v}\), this time the derivitive functions keeps changing and is not a constant. This means that, this time, we really do have to take the integral to get the area and find \(\vec{R}(t)\). Taking the integral on both sides of Equation (3), we get
$$\vec{R}(t)=\int{\vec{v}(t)}dt=\int{\vec{v}_0+t\vec{g}(t)}dt=\int{\vec{v}_0}dt+\int{t\vec{g}(t)}dt$$
$$\vec{R}(t)=\frac{1}{2}\vec{g}(t)t^2+\vec{v}_0t+C.\tag{4}$$
We can find the value of \(C\) in the equation above by doing the same trick as before. Let's set \(t=0\) to get \(\vec{R}(0)=C\). \(\vec{R}(0)\) is the initial position of the object which we'll represent with \(\vec{R}_0\). Substituting for \(C\) into Equation (4), we have
$$\vec{R}(t)= \vec{R}_0+ \vec{v}_0t+\frac{1}{2}\vec{g}(t)t^2.\tag{5}$$
Ok so let's wrap everything up we just did by writing down below Equations (1), (3), and (5):
$$\frac{d\vec{v}}{dt}=\vec{g}=9.8\frac{m}{s}(-\hat{j})$$
$$\vec{v}(t)=\vec{v}_0+\vec{g}(t)t$$
$$\vec{R}(t)=\vec{R}_0+\vec{v}_0t+\frac{1}{2}\vec{g}(t)t^2.\tag{6}$$
Equations (6) are the basic equations of kinematics that we use to describe objects moving near Earth's surface under the action of the Earth's gravity—so they are used for projectile motion problems. By using integral calculus and Newton's laws, we were able to derive the equations of motion for all terrestrial objects—something which alluded some of histories greatest thinkers for the past 2,500 years.
The equations of kinematics governing projectile motion (or you could also call it terrestrial motion) is one of the first great generalizations in the history of physics. One of the major areas of focus on Greg School will be to look at the myriad of applications of each of the great generalizations in physics. We will always, for pedagogic reasons, start off with simple applications, plus a few challenging ones, to get a feel for how they are used in practice. But we'll also always try to head towards the sublime and profound: we'll also focus considerably on some of the most important applications, some of which have greatly transformed our quality of life and our view of the universe.
Standard equations of kinematics
Equations (6) are used to describe the motions of objects moving under the action of gravity near the Earth's surface. But we could be much more general than this and derive the equations of motion for an object being acted upon by any arbitrary force \(\vec{F}\) such that the magnitude of the acelleration \(a\) is constant. To derive some\(^1\) of the equations, we'd go through all the same steps taken to derive Equations (6), except we'd replace \(\vec{g}\) with \(\vec{a}\) to get
$$\vec{v}(t)=\vec{v}_0+\vec{a}t$$
$$\vec{R}(t)=\vec{R}_0+\vec{v}_0t+\frac{1}{2}\vec{a}t^2.\tag{7}$$
Taking the magnitude on each side of both equations in Equations (7), we have
$$v=v_0+at$$
$$R=R_0+v_0t+\frac{1}{2}at^2.\tag{8}$$
There are two other equations which are typically used in kinematics that we'll now work on deriving. The first is pretty simple. We know that if an object is traveling at a constant velocity (where \(\vec{a}=0\)), then its motion can be described by the equation \( Δ R=vt\). Now, the average speed of an object is defined as \(\bar{v}≡\frac{v_0+v}{2}\). But notice that since \(v\) is a constant, it follows that the speed \(v_0\) (at some time \(t_0\)) must be the same as the speed \(v\) (at some time \(t\)). Thus,
$$\bar{v}=\frac{v_0+v}{2} = \frac{v+v}{2}= \frac{2v}{2}=v.$$
In other words, we can substitute \(\frac{v_0+v}{2}\) for \(v\) into the equation \( Δ R=vt\) to get
$$ΔR=(\frac{v_0+v}{2})t.\tag{9}$$
Equation (9) is the third equation in the standard equations of kinematics. To derive the fourth equation of kinematics, we'll just have to do a lot of algebra and make a lot of substitutions. Doing some algebra on both sides of Equation (9), we find that the time is given by
$$t=\frac{2}{v_0+v}ΔR.$$
If we substitute this result into the top equation of Equations (8), we get
$$v=v_0+a\frac{2}{v_0+v}ΔR.\tag{10}$$
Let's multiply both sides of Equation (10) by \(v_0+v\) and do some algebra to get
$$v(v+v_0)=v_0(v+v_0)+2aΔR$$
$$v^2+v_0v=v_0^2+v_0v+2aΔR$$
$$v^2=v_0^2+2aΔR.\tag{11}$$
Writing Equations (7), (9) and (11) below, we have
$$ΔR=(\frac{v_0+v}{2})t.$$
$$v=v_0+at$$
$$R=R_0+v_0t+\frac{1}{2}at^2$$
$$v^2=v_0^2+2aΔR.\tag{12}$$
Equations (12) are the standard equations of kinematics. Which equation is used depends on the initial conditions of the problem which we'll discuss in the next lesson where we talk about the practical applications of Equations (12).
Further Reading
1. “What are the kinematic formulas?”khanacademy.org. Khan Academy. Web. 08/2017.
Notes
1. We'll ignore including the equation \(\frac{d\vec{v}}{dt}=\vec{a}(t)\) since this equation is not included in the four standard equations of kinematics.