Let's apply the kinematics from this chapter to the study of a projectile. A projectile is NOT just something that flies. It is a somewhat fictitious object in the sense that only the effect of gravity is considered. It is assumed that the object experiences no air drag from the atmosphere, no thrust from rockets or propellers, no lift forces from wings or rotational motion. [br][br]In this sense, the closest analogs in real life are massive objects that are thrown or launched at modest speeds. Maybe throwing a rock, putting a shot (track and field) or a basketball being shot, albeit maybe not a very long range shot. In all these cases, while air drag IS present, it is fairly tiny as compared with the gravitational force, so their trajectories are not affected much by the presence of the atmosphere.[br][br]What we can say about a projectile from a physics standpoint is that [b]it has a downward acceleration equal to that due to gravity, a[sub]g[/sub][/b]. We endow the object with an initial velocity that must be known to predict the path the object follows. We further must define the coordinate system to use - both the location of the origin and the orientation of the axes. The simple thing to do in that regard is to [b]place the origin at the location of launch and to set the x-axis in the horizontal direction and y-axis vertically upward[/b]. Recall that the best choice of coordinates is to make one axis parallel to the acceleration, which is downward in this problem. Putting the origin at the starting location makes the initial position a zero vector. We have no need of a third z-axis since all standard projectile motion on earth is planar motion so long as we don't worry about the fact that the earth is rotating on its axis... and we won't right now, but we will later on in this course.[br][br]Mathematically, from the information given above, we have the following three facts:[br][br][center][math][br]\vec{r}_0=\vec{0} \\[br]\vec{v}_0=v_{x0}\hat{i}+v_{y0}\hat{j}\\[br]\vec{a}=0\hat{i}-a_g\hat{j}.[br][/math][/center][br]This is really all we need to find out everything we want to know about the projectile. Naturally, if you like numbers, we need to know the numerical values of the velocity components and the acceleration is of course approximately 10m/s[sup]2[/sup]. The only meaningful things to know about this projectile are how it moves (velocity) at any time after launch, and where it is located (position) any time after launch. Both of those will be time-dependent vectors, and if you paid attention in your reading of this chapter, you know how to arrive at them.[br][br]To get the velocity at some time after launch, t, we need two things: The initial velocity and the change in velocity between launch (t=0) and time t. The change is arrived at by integrating the acceleration vector. Let's do this now.[br]

STEP 1: Integrate the acceleration to find the change in velocity.[br][br][center][math]\Delta\vec{v}=\int_0^t \vec{a}\;dt \\[br]\Delta\vec{v}=\int_0^t (0\hat{i}-a_g\hat{j})\;dt \\[br]\Delta\vec{v}=-a_gt\hat{j}.[br][/math][/center][br][br]STEP 2: Find the velocity by adding the change to the initial velocity. This will be valid for any time 't' that is entered into the equation.[br][br][center][math]\vec{v}=\vec{v}_0+\Delta\vec{v} \\[br]\vec{v}=v_{0x}\hat{i}+v_{0y}\hat{j} -a_gt\hat{j} \\[br]\vec{v}=v_{0x}\hat{i}+(v_{0y} -a_gt)\hat{j}.[br][/math][/center][br]The speed is naturally the magnitude of this velocity vector. Notice that while the x-component doesn't change in time, the y-component does. The speed will be minimum when the y-component is zero. This occurs when the projectile is at the top of its trajectory. This must be the case since the velocity is going purely in the x-direction - meaning it is not going upward (rising) or downward (falling) at this moment in time.[br][br]STEP 3: To find out how the projectile has been displaced at any time 't', we integrate the velocity to find the displacement vector. This is how it has moved with respect to its starting position. [br][br][center][math]\Delta\vec{r}=\int_0^t \vec{v}\;dt \\[br]\Delta\vec{r}=\int_0^t v_{0x}\hat{i}+(v_{0y} -a_gt)\hat{j}\;dt \\[br]\Delta\vec{r}= v_{0x}t\hat{i}+(v_{0y}t -a_gt^2/2)\hat{j}.\\[br][/math][/center][br]STEP 4: Add the displacement to the initial position so that we know where the object is at any time 't'.[br][br][center][math][br]\vec{r}=\vec{r}_0+\Delta\vec{r}. \\[br]\vec{r}=r_{0x}\hat{i}+r_{0y}\hat{j}+v_{0x}t\hat{i}+(v_{0y}t -a_gt^2/2)\hat{j} \\[br]\vec{r}=(r_{0x}+v_{0x}t)\hat{i}+(r_{0y}+v_{0y}t -a_gt^2/2)\hat{j}. \\[br][/math][/center][br][br]Beyond this, the only thing to know is that the components of the initial velocity are just found using the trigonometry of triangles - cosine for the x-component and sine for the y-component.

LIMITATIONS: The work above amounts to finding where the projectile is (position) and how it's moving (velocity) for all times 't' after launch. [b]There is no assumed landing or impact with the ground in the calculations[/b], so if we insert a very large value for time, we will find the projectile far to the right and much lower than the starting location as if the object had been shot into something like the grand canyon in which it can fall unimpeded for a long while. [br][br]I should also remind you that there is no air drag in the equations above, so these results do not apply well to faster-moving projectiles on earth or in any environment with an atmosphere. [br][br]Even in an air-less world, the equations above do not account for the curvature of the planet, so trying to use a very fast projectile in order to set something in orbit (since that's what an orbiting object really is) will not be handled correctly. [br][br][u]Time of Flight[br][/u]We can find the time of flight for a projectile if we know the landing height. A common case is to assume the landing height is the same as the take-off height (which we called zero for convenience). If we take the equation for the position above and solve for the y-component to equal zero, we get two values. One value is the trivial solution that at t=0 we are on the ground, and the other is the one that tells us how long it takes to be back on the ground. This is the time of flight. Let's do the math:[br][br][center][math][br]r_y=r_{0y}+v_{0y}t -a_gt^2/2 \\[br]0=0+v_{0y}t -a_gt^2/2 \\[br]t=0,\frac{2v_{0y}}{a_g} \\[br]t_{flight}=\frac{2v_{0y}}{a_g}.[br][/math][/center][br]Here, [math]v_{0y}[/math] is just the y-component of the launch velocity vector. In terms of angle of launch (above horizontal) and speed of launch, this is [math]v_{0y}=v_0\sin \theta.[/math][br][br][u]Range of Projectile[br][/u]If we know the time of flight, it is easy to find the range of the projectile, or how far it lands from the launch position. Keep in mind that this is not the distance it travels during flight. It follows a parabolic trajectory during flight which will be longer than the straight-line distance from launch position to landing position. [br][br]To find the range, we really just want the x-component of the position vector at the moment it lands. Since we start from the origin (our choice, for convenience) this x-component must be the range. We already know the time corresponding to landing which we called time of flight above. Plugging that time into the x-component of the position gives us:[br][center][math][br]r_x = r_{0x}+v_{0x}t \\[br]r_x = 0+v_{0x}t \\[br]r_x = v_{0x}\frac{2v_{0y}}{a_g} \\[br]r_x = v_{0x}\frac{2v_{0}\sin\theta}{a_g} \\[br]r_x = v_{0} \cos \theta \frac{2v_{0}\sin\theta}{a_g} \\ [br]r_x = \frac{v_{0}^2}{a_g} 2 \cos \theta \sin \theta \\[br]\text{Using the double angle formula gives} \\[br]r_x = \frac{v_{0}^2}{a_g}\sin 2 \theta .[br][/math][/center][br]This result suggests several truths about drag-free projectiles. First of all, the range is sensitive to choice of initial speed. While instinct might incorrectly suggest that doubling the initial speed will double the range, the reality is that [b]the range will be quadrupled if the initial speed is doubled, since the speed term is squared[/b]. The act of shooting free throws in basketball would be easier if the term wasn't squared since the range wouldn't be as sensitive to little errors in launch speed when we take a shot.[br][br]This is also an opportunity to think about the meaning of functions like in the chapter on mathematics. We should ask ourselves why the speed term is squared. There must be two influences that it has on the range. It turns out that doubling the launch speed will double both how long the object is in the air (flight time) and it will separately double the rate at which it moves along in the horizontal direction. These two independent factors make the speed squared in the range equation. Keep in mind that you could have inspected the math to find the reasons for yourself just by going back through the steps.[br][br]The other thing the range equation tells us is that with a fixed launch speed, the maximum range is the one that maximizes the [math]\sin 2\theta[/math] term. Since 90 degrees inside the sine function maximizes its value,[b] a launch angle of 45 degrees (so that 2(45)=90) maximizes the range[/b]. Your instinct probably tells you as much. Keep in mind that sports with air drag do not work this way. Long passes in football and tee shots in golf go farther at smaller angles than 45 degrees. Our result above only applies to drag-free and spin-free projectiles.[br][br]Due to the symmetry of the sine function around 90 degrees - that sin(80[sup]o[/sup])=sin(100[sup]o[/sup]), etc - we can see that [b]shooting a projectile above or below 45 degrees by the same amount will lead to the same range[/b]. In other words, the math tells us that a projectile shot at 40 degrees will fly as far as one shot at 50 degrees, or that one shot at 30 degrees will fly the same distance as one shot at 60 degrees (both 15 away from 45). What will not be the same for those two scenarios is the time of flight. The math and the animation show that the steeper launch angle will have a longer flight time.