Finally, the core of kinematics. This is what many people consider to be the most useful application of kinematics. If you remember studying vectors, it will get very, very handy right now. Remember that vectors can be defined with a magnitude and angle from the positive x-axis, and so we can launch a projectile at with an initial velocity, at a certain angle, and it will yield this:
Figure 1: Different modes (examples) of projectile motion Ah yes, the beautiful, projectile motion! You may have noticed that all objects you throw will follow this curve. This curve is called a parabola, which is a fancy math term that you probably don't really need to worry about. We will use this term from time to time to refer to the nature of the trajectory, or flight path the object follows. Now, instead of a static picture, we'll put an interactive demo to show how this kind of motion plays out in real time.
This is what many people consider to be the most useful application of kinematics, and many interesting formulas and problems arise from this as a result. Remember that vectors can be defined with a magnitude and angle from the positive x-axis, and so we can launch a projectile at with an initial velocity $v_0$, at a certain angle $\theta$, and it will yield this: Figure 1: Different modes of projectile motion Now that you've seen our diagram, it's time to experience it. Or at least, experience it the best you can in an online setting. We have made a projectile motion demonstration with adjustable parameters for launch speed and angle. Do note that the projectile will bounce because we thought it would look better, but the thing we're concerned with here is the initial motion before the first bounce. Anyway, enough said. Here's a demo so you can get a feel for projectile motion:
Two-Dimensional Motion Demo
Input the velocity and angle from horizontal:
at degrees
Look away from the demo for now and back at the picture. You may have noticed that in the drawing, there are two projectiles that are launched at completely different angles yield the same range. The reason for this is more mathematically intense than we can handle, but conceptually, all you need to know is that if you shoot an object at an angle, say $10$ degrees, then if you shoot it at $90$ minus that angle (in degrees), it will also land at the exact same range.
So for $10\degree$, the angle $80\degree$ will yield the same range. Now, just by doing basic arithmetic, we can see that $45\degree$ does not actually have another counterpart angle! What does this mean? Launching it at $45\degree$ will give you the maximum range a projectile can go (still neglecting air resistance, in real life, this is more like $22\degree$, since air resistance exists).
However, this actually only holds for when your change in vertical displacement is zero. So if I threw a rock off a cliff, the maximum angle of range without air resistance is NOT $45\degree$. If you are curious to know, it's somewhere around $35.3\degree$. But that's going off topic. There are other important concepts to know and realize about projectile motion that await! Most likely, you have seen or are seeing the parabolic trajectory that the object follows. This has a very simple proof and will be explained later on. First, if you remember the formulas for total flight time, and the maximum height, you might be wondering: Can we derive some form of them here as well? Well, yes!
Calculations with 2d Motion
Actually, the idea behind working with kinematics in two dimensions is the fact that you can simply split up $v_0$ into its horizontal ($x$) and vertical ($y$) components, and we realize that the time for the horizontal component to travel the maximum range is the same time it takes for the object's vertical component to travel up and down. They are linked by a single quantity, time, but otherwise you can treat them as separate in their own respect.
IMPORTANT! This concept (above) is the key to working in kinematics in 2 dimensions!
Then, we recognize that velocity is a vector that can be split into its components, $v_{0x}=v_0\cos \theta$ and $v_{0y}= v_0\sin \theta$. Considering the accelerations along each axis, we see that $v_{0x}=v_{fx}$ because the horizontal velocity is constant, since there is no acceleration in the horizontal direction (neglecting air friction, like usual). In the $y$ direction, the acceleration $a_y$ is equal to $-g$. Thus, we can find the maximum height and time of flight easily by replacing $v_y$ with $v_0\sin\theta$, which is its equivalent in our case.
$$t_{total}=\frac{2v_{0y}}{g}=\frac{2v_0\sin \theta}{g}$$ $$H=\frac{v_{0y}^2}{2g}=\frac{v_0^2\sin^{2}\theta}{2g}$$ However, since this motion is horizontal as well, there is one more quantity that is of use: the range of the projectile. To find this, we first realize that the horizontal velocity is constant, and that we simply need to compute:
$$R=v_{0x}t_{total}$$ We know the horizontal velocity and total time, so we substitute that in:
$$R=v_0\cos\theta \cdot \frac{2v_0\sin\theta}{g}=\frac{2v_0^2\sin\theta\cos\theta}{g}=\frac{v_0^2\sin{2\theta}}{g}$$ ...Since $\sin{2\theta}=2\sin\theta\cos\theta$. We recognize that this is a maximum for a particular $v_0$ when $v_0 = 45\degree$. When this angle is achieved, the equation simplifies to just $R=\frac{v_0^2}{g}$
Additionally, using calculus, we can prove this is the case:
Use the chain rule here:
$$\frac{d}{d\theta}(\sin{2\theta})=2\cos{2\theta}$$ Setting this to zero (to perform the first derivative test) yields:
$$2\cos{2\theta}=0$$ $$\cos{2\theta}=0$$ We recognize that the solution for $0\degree≤\theta≤90\degree$ is $45\degree$!
The next page will prove why the trajectory is parabolic.
2d Motion Concepts
There are many cool concepts about projectile motion. Firstly, if you notice, there are always going to be two solutions of range, for any given height. For example, if you draw a straight line at any height less than the maximum, you'll notice that it intersects two points on the parabola. This is a basic concept that is important to notice and realize. However, there is only one solution for the maximum height.
Another really cool thing is that remember when we said it takes half the time to reach the peak and half to come back down? The same principles apply. So if we know that's true for the vertical velocity, and if we add a horizontal velocity that is constant throughout the trajectory (assuming no air resistance), the same rule applies: the object thrown at some velocity $v$ will impact the ground with the same magnitude!
If you are more interested in the calculations part, you can take a look at our Algebra-based level. But this is a more useful concept for calculating the grand scheme of things, so you might as well. You might be thinking that modeling the object's trajectory must be pretty hard to do.
However, we don't have to model this with a graph. We can actually split up the velocity vector into its $x$ and $y$ components! And you know what else? These components can be treated individually. The horizontal velocity can be used completely separate from the vertical velocity. The key thing to note here is that these two components are linked by the quantity time, because if you think about it, the time that it takes the horizontal component to reach the final range is the same time it takes the vertical component to travel up and down, as if it were on its own.
Conceptually, this is pretty much all there is to projectile motion. Next up is a short and brief lesson that will talk you through some things that will be helpful to know, particularly for unit 2.
The Trajectory Formula
In order to prove that the trajectory is parabolic, we need to relate x and y. To do this, we write general equations for each like this:
$$y=v_{0y}t-\frac 1 2 gt^2$$ $$x=v_xt$$ Then, we solve for t to get:
$$t=\frac{x}{v_x}$$ Plugging this into the equation for y, we get:
$$y=\frac{v_{0y}x}{v_x}-\frac{gx^2}{2v_x^2}$$ This is a quadratic equation for $y$ in terms of $x$, which thus proves the trajectory is parabolic! We can actually simplify this further. Below is the steps, although some are omitted for the sake of simplicity (and so that it doesn't take up the entire screen).
$$y=\frac{v_0\sin\theta \cdot x}{v_0\cos\theta}-\frac{gx^2}{2v_x^2}=x\tan\theta-\frac{gx^2}{2v_0^2cos^{2}\theta}$$ This is known as the trajectory equation, and you can use it to calculate the height $y$ for a given displacement $x$ it has traveled of its maximum range and vice versa. However, it is important to note, as with a parabola, there are two points in the range where the height is the same. If the problem only asks for one of them, use your best judgement and context to decide which is the answer. Having seen this, let's work on a problem together:
Jebron Lames is shooting hoops. He throws the basketball at an angle of $60\degree$ with the horizontal with an initial velocity of $5~\textrm m /\textrm s$ towards a hoop that is $0.5~\textrm m$ taller than he is. How far away is he from the hoop, if he made a perfect swish on that shot? (Hint: The ball has to enter the hoop from above.)
Figure 2: He shoots (he calculates), he scores! We can use the trajectory formula we just derived to solve this problem. We can use the quadratic formula to obtain a general solution for $x$:
$$\theta=60\degree$$ $$v_0= 5~\textrm m /\textrm s$$ $$y = 0.5~\textrm m$$ $$y=x\tan\theta-\frac{gx^2}{2v_0^2cos^{2}\theta}$$ Using the quadratic formula gives us a huge mess, but it can be simplified to look much nicer. The algebra in between is omitted for the sake of clarity. It is up to you to do the math. Challenge yourself and get through the algebra!
$$x=\frac{\tan\theta+\sqrt{\tan^{2}\theta-\frac{2gy}{v_0^2}}}{\frac{g}{v_0^2\cos^{2}\theta}}=\frac{v_0^2\sin\theta\cos\theta+v_0^2\cos\theta\sqrt{\sin^{2}\theta-\frac{2gy}{v_0^2}}}{g}$$ Now you might say the actual correct way to do it is using "$±$", but Jebron knows there is only one possible answer if the ball is actually to go through the hoop. Eric was nice enough to give you a hint, the ball needs to enter the hoop from above. That means, the ball's vertical motion is downwards, which means it is farther in the parabolic trajectory. Since it is farther, we must use only "$+$". Also, as an aside, the maximum range equation can be simplified from this, as the range equation is simply when $y=0$, so the square root term is reduced.
Eric side note here. The term where use the $-$ is actually a valid solution as well. The ball will pass through the same height twice during its motion (think of what a parabola looks like) but it won't go through the hoop if it's moving upwards, which is what the smaller solution represents. Enough rambling. Now if we plug in the values, we get:
A ball is thrown an some initial velocity at an angle of $52\degree$ above the horizontal. It reaches a max height of $7.5~\textrm m$. How high would it go if it was instead launched directly upwards with the same magnitude of initial velocity?
To solve, we can use our knowledge of the maximum height in trajectory:
$H_1=\frac{v_0^2sin^{2}\theta}{2g}$ Solving for $v_0$, we obtain: $v_0=\sqrt{\frac{2gH_1}{sin^{2}\theta}}$
If it was thrown straight upwards, the max height equation reduces to $H_2=\frac{v_0^2}{2g}$
Remember, the key to solving problems like these is to identify your known values and map out your problem solving strategy, then solve the equations algebraically, and then finally, last of all, plug in your values and obtain the final answer. Many physics problems in the future will not be as simple and straightforward as plugging in values, which is why you should try to get in the habit of solving problems like this. Next up is a short and brief lesson that will prepare you for your next unit!
The AntiMatter Lab 
