 the pitch of a softball into the air, free to experience only gravity as it moves, the strike of the bat and the return of the ball to gravity's embrace, the leap of the basketball player into the air in anticipation of dunking the ball. This is a kind of poetry written in the air by gravity, the satisfaction of launching a football into the hands of gravity to have it delivered exactly to the place where it needed to be. All of these are examples of projectile motion here written out in a kind of concocted perfection as the ball is launched from a moving cart and the ball returns to the cart because the horizontal component of its motion is not affected by the fact that it has been granted a vertical component of motion. Let us explore this kind of motion and all of the benefits and mathematical beauty that it reveals in nature. Let's begin to look more deeply at projectile motion and build a toolkit based on our previous explorations of motion in more than one dimension to describe this motion. First, let's lay out some of the key ideas that we're going to cover in this lecture. First of all, we will find that projectile motion is a special kind of motion that happens when motion in one dimension is unaffected by motion in any other dimension. The motion of airborne objects near the surface of the earth is an excellent example of projectile motion. We can describe that kind of motion using the ideas we have already developed. And in cases of projectile motion where there is constant acceleration along only one of the coordinate directions, the observed trajectory in space will be a parabola. These are the ideas that we will develop more fully in this lecture. Now, there are, as you've seen, many common examples of projectile motion that we experience every day. Consider the cartoon over on the right which show silhouetted figures in various games of sports involving catching, throwing, or deflecting a ball. A ball flying through the air, moving only under the force of gravity, or a person, after pushing off the ground in a jumping maneuver, now unable to contact any other surface to exert any other forces and thus subjected only to gravity. These are examples of projectile motion that are very common in everyday life. Let's consider a very specific situation which will illustrate all the features that we need to explore in projectile motion. Consider the situation that is depicted on the right hand side of this slide. We have some kind of projectile launching device. In this case it's a tube that can allow air to force a projectile out of one end of the tube. This is a blowgun or an air gun and it has been aimed so that the barrel points straight at this metal can which is hanging here from a stand. When the ball leaves the barrel of the gun, the can is simultaneously dropped from the stand. Both the can and the steel ball are now executing projectile motion. The question that we would like to explore is will the steel ball hit the can or will it miss the can? We can analyze the situations, for instance, such as firing a steel ball with an initial velocity and dropping a can from rest simultaneously using the equations of motion that we have been playing with and developing so far and try to answer this question. But before we do that in a mathematical sense, let's take a look experimentally at what would be required to set up exactly the situation that I have just described. The essence of this demonstration are three pieces. One is a metal tube that is capable of holding a small ball bearing. This metal ball bearing is like a bullet that one can load into the barrel of the tube in the front here. The tube is connected at the back to a hose that goes to a supply of air that's provided in the classroom workbench environment. So I'm able to turn the handle on the air nozzle and get it to force air through the tube and push on the ball bearing, eventually shoving it out of the end of the tube. So we can already see that there is a force that's present to accelerate and launch the ball, and that is the force of the air pushing on the ball inside the tube. But once the ball exits the tube, the air is really no longer acting on it. That force was only present to accelerate the ball inside the tube, and it diminishes quite rapidly once the ball leaves the tube. So out here, there is no air pressure pushing from one direction only on the ball, and so we would expect that there would be no forces acting on the ball. But of course, that's not true. There is a force that will act on the ball. It's acting on the ball constantly throughout this, and it really begins to take over once the ball exits the tube, and that is gravity. Gravity will tend to want to accelerate the ball down at a rate of about 9.8 meters per second squared. So once the ball exits the tube, we can think of at least one force that's definitely acting all the time on this ball, and that is gravity, and it should be exerting a force in the downward direction, yielding an acceleration of 9.8 meters per second squared. Now this ball is made of metal, and as a result of that, it has some inertia that prevents it from falling prey easily to small forces that would tend to accelerate it. Now as we know in the real world, there's air, and if we wave our hands around, we can feel the resistance of air. And in fact, air resistance is a force that always tends to sap motion out of an object because it always resists the motion of an object in its direction of motion against that direction. So certainly, air resistance will be acting on this ball, but it turns out to be a relatively minor force compared to gravity, and so we will neglect it in this particular experiment. Let's look at some of the other elements of this. Over here, I have a small electromagnet. This is a device that only produces a magnetic field when an electric current is run through the device. And if you look closely here, you'll see that there is a white pair of wires that goes up and into the electromagnet. Now I can power it using this device over here, which is a power supply. It's simply connected into the wall, and it converts the voltage from the wall into something that's a bit more friendly for this electromagnet. And so with a flick of the dial here and a gentle turn, I can start running a current through this by exerting a voltage of about six volts on the wire. So now that I have a voltage, and thus the potential for a current inside of this device, there should be a magnetic field. And I can verify this by switching off the power supply and showing you that I'm unable to get this metal can to stick to the electromagnet. On the other hand, if I switch on the voltage, and now I attempt to stick the can to the magnet, bingo! It easily remains there in place, neither moving up nor down nor left nor right, nor forward nor backward. There are no net forces acting on this can, and it is now in an equilibrium state. It cannot move at all. So let's think about what we've done. We have a device that can accelerate a small metal projectile, a small metal ball bearing. That projectile conveniently is aimed when it leaves this barrel at the can. I took the effort to line up the barrel of this to point, like a little rifle, at the can. Now here's the last ingredient in this exercise, and that is that the current that's powering that electromagnet is controlled by a little switch right here. The current is on, the metal can is clearly attached to the electromagnet. But if I push the switch and I break the electric circuit, I interrupt the flow of current to the electromagnet, and what happens? Well, an electromagnet is only magnetic when there's a current flowing through there. If I interrupt the electric current, there will be no magnetic field. Without the counterbalancing magnetic force from the electromagnet, gravity is now free to act solely and only on this can, pulling it closer to the center of the earth. So here's our experiment. Now that we've had a chance to see how such an experiment could be conducted, let's begin to analyze this using the principles of motion and physics that we have developed so far. We should begin by describing the situation using a coordinate system with an x and a y-axis, and I have denoted the units of distance along each of those axes to be meters. We're going to use this to represent and describe the steel ball, the can, and the features of the motion. We can then place a mark in the coordinate system, indicating the initial location of the steel ball, the moment that it exits the blowgun, and is no longer being acted upon by the compressed air inside the tube. And we mark this location using a blue dot shown over here on the left in the graph. Note that I have placed the blue dot at one meter above the horizontal, and along the horizontal axis, it actually sits at zero meters. This point one indicates the leftmost extent of the x-axis here. This location under the blue dot is actually zero meters in the coordinate system. Let us then also mark the initial location of the can just at the moment that it is released from the stand that it is dropped. Remember, the exiting of the steel ball from the gun and the dropping of the can occur simultaneously, and the moment in time that we will state where those events both occur is t0, t equals zero seconds. That's our initial time for the motions to begin. We mark the location of the can with a red dot. Please note that I have placed the can at a height of three meters above the floor, and two meters from where the steel ball exits the gun. And as we are going to see, the motion of the steel ball will look like the black path, the curve that I've now drawn here, shown in the graph. Before we begin to actually explore the motion where this path emerges from in a mathematical sense, let us explore key points of the motion, accepting this curve as the correct mathematical description, before we dive into the details I just mentioned. There are a few key and interesting points along the motion curve of the steel ball projectile. The blue dot already marks one of them. It's the initial location of the steel ball in x and y. We can denote that location x0 comma y0. It is here that the steel ball leaves the blowgun with an initial velocity, and I have represented the arrow of that velocity vector, and written it out mathematically here. It has a horizontal component. It has a vertical component. And one of the key features of this velocity is that the barrel of the gun, and thus the steel ball when it exits the barrel, are initially aimed at where the can begins its motion. So where the can is hanging from the stand in the experiment. So note that the ball begins at 0 comma 1 meters. The can is at 2 comma 3 meters. And so if we turn off gravity for a moment, if we want the steel ball to hit the can, its velocity vector has to point along a line that executes equal motion in the x direction as in the y direction in the same amount of time. So after some amount of time, we want the steel ball to be here if there's no gravity. Thus the can can't fall. It just sits where it's parked when we initially fire the ball. And then in the next unit of time, we want the steel ball to traverse this distance continuing along the diagonal between these grid points to strike the can. The initial velocity of the ball must be aimed directly at the can, even though gravity will act on both of these objects once they are released from either the stand in the case of the can, or from the tube in the case of the steel ball. So we will need a velocity vector in this depiction of the situation whose horizontal component is the same as its vertical component, regardless of what the total magnitude of the velocity is. We just, based on this picture, need the components to be equal. Now in a different situation where the coordinates of the can are not what I've shown here, we might need a different kind of breakdown of the components of velocity. And these are things that we can explore in homework and in class. So this is one interesting point in the motion, the initial location of the steel ball with its initial velocity v0. The apex of the motion, that is, the highest point along the motion curve in y, that the steel ball manages to reach, is another interesting key point in the motion. Since gravity accelerates the steel ball downward toward the surface of the earth, there will come a point where the ball no longer moves up or down its vertical motion appears to pause for a moment. And then after that point it reverses its vertical motion and begins to fall from rest at the top of the motion curve. At this point the ball is only moving horizontally. There are no forces acting along the horizontal direction. The ball continues to move horizontally the whole time, but its vertical motion is more complex because at first it's going against the direction of acceleration due to gravity, but eventually gravity wins and it reverses the direction vertically of motion of the ball. The horizontal component, however, is independent from the vertical component. This will allow us to analyze this motion. This purple arrow indicates the velocity of the ball, which only has a horizontal component at the apex of the motion. As you can see, the motion has two halves before the apex and after the apex. Before the apex, the combination of horizontal and vertical motion for the steel ball points up and to the right, but after the apex the combination of horizontal and vertical motion points down and to the right. And this combination is ultimately what produces this curve, whose actual mathematical shape is a parabola. Let's begin to analyze the motion of the steel ball and consider what's happening, but for now only in the horizontal direction. The initial velocity in the horizontal direction is given by the x component of the velocity v0x and it points in the i hat direction in the positive x direction and this I have written here as a unit vector indicating just the direction of the horizontal component along x. Because there is no source of acceleration in the horizontal direction, the steel ball has left the blowgun. It's no longer being pushed on by the air inside the gun. There is no force acting horizontally if we neglect air resistance and we are doing that in this problem as an aside for a small steel ball. Neglecting air resistance is a reasonable assumption. It has a fairly aerodynamic shape. It's also heavy and not so subject to the whims of air resistance as it moves through the air. So we can approximate that there is no source of acceleration or deceleration in the horizontal direction and so after it exits the blowgun and the blowgun can no longer act on the steel ball, there is no force that can speed up or slow down the component of velocity in the x direction. Remember the equation of motion, that the velocity at a later time is equal to the velocity at an earlier time plus the acceleration along that direction times the time. But since the component of the acceleration in the x direction is zero, we see that only in the horizontal direction the velocity remains constant throughout the entire motion and in fact at every point in the motion until an external force acts on the ball, for instance the steel ball strikes the floor and the carpet and the floor change what the ball is doing the horizontal velocity will always be the same. So we actually now have an equation of motion that gives the x coordinate at any time t after the steel ball is fired from the barrel of the blowgun and again to remind you it begins its journey at 0 comma 1 meters and the can begins its journey, although we're not talking about the can right now at 2 comma 3 meters and here is the equation of motion for the steel ball in the x direction. This will tell you the x coordinate of the steel ball at any time t after being launched from x0 with a velocity v0x. So this isn't so bad, it's nice not to have acceleration in one of the directions, it makes the equation fairly simple. Now let's consider what's happening in the vertical direction. The initial velocity in the vertical direction has a magnitude v0y and it points originally in the positive y direction. The steel ball is moving to the right and up initially, it's being fired upward toward the can. So its initial velocity is to the right and up and so it must be that the vertical component of its overall velocity vector points upward with some initial length v0y at time 0 and that's denoted by the blue arrow in the graph. Now gravity of course is acting in the vertical direction, it exerts a constant downward acceleration of 9.8 meters per second squared in magnitude and thus the acceleration component in the y direction is negative g where g is 9.81 meters per second squared. This will cause the vertical velocity to change with time. Again remember the equation of motion relating velocity acceleration and time, the velocity at any time t is given by the initial velocity plus the acceleration times time but in the vertical direction the acceleration is negative g times time. So we have this equation which tells us that the velocity will while first being positive eventually be overtaken by the acceleration due to gravity reversed and point downward more and more and more as gravity accelerates the ball downward toward the floor. Now after the apex and remember that the apex is where the vertical velocity is exactly zero meters per second in the vertical direction the steel ball is neither moving upward nor downward at the apex. After that point acceleration reverses the direction of velocity and the velocity vector points down and its magnitude will grow larger and larger as gravity continues to act on the ball. So we can write an equation of motion that gives the y-coordinate at any time t and that's shown here. The y-coordinate at any time t will be its initial y-coordinate plus its initial velocity in the y-direction times time minus one-half gt squared and the minus sign is because the acceleration is in the negative direction so a in the y-direction is replaced by negative g. So let us now write the two equations one for the horizontal location and one for the vertical location at any time t. I've written those out here on the slide and I've now indicated here this expected trajectory for the motion and we'll see again where this arises from the moment although we already have a kind of qualitative understanding of why this motion should be curved as the horizontal component moves with constant velocity and no acceleration whereas the vertical component has a changing velocity due to the constant acceleration due to gravity. Now these are the equations that tell us position in space given velocity and acceleration and we can similarly write the components of the velocity itself at any time t and again we have a horizontal component that is constant with all time because no acceleration acts on the horizontal component or neglecting air resistance. In the vertical direction however the velocity components magnitude does change with time and in fact at at some point given the acceleration it will reverse and point downward. Now using the equations on the previous slide we can return to the original question that motivated all of this mathematical and physical activity. Will the steel ball so fired strike the can? To hit the can the steel ball must reach the same coordinates as the can at the same moment in time t. Now we know what the x coordinate of the can will be at any time t because the can is dropped from rest. It's not moving to the left it's not moving to the right when it's dropped and so its x coordinate is the same throughout all of its motion. x sub can and we'll use the subscript can to denote quantities for the can is the same as x zero can the initial position of the can along the x axis. So we can ask a question at what time does the steel ball reach an x value of x sub can? Again taking a look at this picture the x location of the can is two meters from where the steel ball started which is considered to be zero meters along the x axis. So we can actually use algebra to answer the question. We have an equation of motion for the x coordinate of the steel ball at any time t after it's launched. We know that we want to find out at what time t x becomes x sub can which is two meters. So we plug in x sub can set that equal to the initial position of the steel ball plus v zero x times t. Well we know all of these quantities except the time. We know x can it's two meters. We know x zero for the steel ball it's zero meters. We know v zero x for the steel ball it's three meters per second which was written on a previous slide. We're assuming that the steel ball is fired with an x initial velocity of three meters per second and a y component of velocity that's also three meters per second in the upward direction. So v zero x is three meters per second and if we plug all these numbers in and rearrange and solve for time we find out that the time at which the steel ball reaches an x coordinate of two meters along the x axis which is where the can is is two thirds of a second or point six six six dot dot dot seconds after the steel ball is launched. Now using that time we can take the t value of two-thirds of a second and we can plug it back into the equation for the y coordinate and find out what the y coordinate of the steel ball will be when it reaches that x coordinate. And we find when we do this and I encourage you to try this yourself go back a slide get those equations of motion for x and y v zero x v zero y g and t. We now know the time from the x equation we can substitute that into the y equation and solve for the y coordinate at t equals two-thirds seconds when x equals x can two meters. And when we do that we found out that the y coordinate is point eight two meters when t equals two-thirds of a second. So we know where the steel ball is supposed to be when it reaches the x coordinate of the can. It will be at a height of point eight two meters at two-thirds of a second after its initial launch. So let's think about the can. Where will the can be at t equals two-thirds of a second? Well the can is merely being dropped at the very same moment that the steel ball leaves the blowgun. We call that time t zero and in our problem that's zero seconds. At first as gravity begins to act with constant acceleration on the can its initial velocity is zero. It has no velocity in the x direction. It has no velocity in the y direction. But as gravity continues to act with constant acceleration on the can and since gravity is the only force that acts on the can the can begins to accelerate straight down along the vertical axis toward the ground and that acceleration in the y direction is negative g there is no acceleration in the x direction so there will never be any velocity in the x direction. Now this is the same constant acceleration that the steel ball is experiencing. The can's velocity continues to increase more and more but it only ever has a vertical component since no forces act in the horizontal direction and there was no initial velocity imparted by any force to the can when it was released in that direction. The position and velocity of the can are then given by these equations of motion. The horizontal position of the can is the same at any time t. It's always x zero can which is two meters. The y position of the can however is a bit more complicated. It is the initial position in y of the can which is three meters above the ground plus the initial velocity of the can times t well the initial velocity is zero it starts from rest so in fact this term goes away because zero times t is zero and then we have to add to that one half a t squared well the acceleration is negative g so we have negative one half g t squared and by throwing out this middle term because it's zero we can simplify this equation to y zero can minus one half g t squared and the velocity of the can is given only by a y term there is no x term because it has no velocity in the x direction to begin with there are no accelerating forces in the x direction so there is never any x component of the velocity there's only a y component of the velocity and it's given by v zero y can minus g t because the can is accelerated down from rest and in fact since v zero y can is zero this is just negative g t in the j hat direction. So will the steel ball hit the can in other words after t equals two-thirds of a second so two-thirds of a second after the can is dropped where will the can be in height well we know where it will be in the horizontal direction it will be exactly where it started it will be at two meters what will its y coordinate be well it started at three meters and if we plug in two-thirds of a second and solve given its initial height the acceleration and the time which is two-thirds of a second we find that the y coordinate the height of the can above the ground after two-thirds of a second is point eight two meters and if this looks familiar there's a reason why the steel balls coordinates that t equals two-thirds of a second are x equals two meters and y equals point eight two meters and the cans coordinates at the same time two-thirds of a second are also two meters and point eight two meters in the x and y directions respectively. So we conclude that the steel ball hits the can but how can this be if we think about the initial situations of the steel ball being fired upward toward the can and the can being dropped from rest it seems amazing that in fact the ball hits the can but the reason for this is because the steel ball was initially aimed at where the can started and after they both enter projectile motion they both experience the same constant acceleration and so projectile motion is extremely powerful it is motivated simply by the fact that we have independent motion in these two directions x and y we could easily extend this to a third direction z if we wanted to and in the special case of objects in projectile motion near the surface of the earth we have this constant downward acceleration from gravity and so we find that we have these situations where we can not only describe very well the motion using the mathematics we've already developed but for this specific situation where a projectile is aimed at a target and the target and the projectile both enter projectile motion in the gravitational field of the earth with no other forces acting on them at the same time that in fact they meet at some point in space later it's really remarkable now there are many examples of projectile motion in the universe a basketball free throw where you carefully aim and then throw a basketball up above a hoop in the hopes of getting it to come down and enter the hoop and make a basket throwing any ball in fact into the air in such a way the the ball will observe projectile motion once it leaves your hand because it will have no forces again neglecting air resistance it will have no forces acting in a direction perpendicular to gravity so along the horizontal and it will feel only the constant acceleration due to gravity in the vertical direction another example is jumping while moving if you are running and then jump into the air you now no longer have any surfaces to touch you can't change your motion and you will be in projectile motion in the gravitational field of the earth so think about track and field long jumping things like that jumping over hurdles those kinds of events involve ideally projectile motion where the body makes no contact with another surface until the motion is completed archery riflery any activity where an object is launched into the air and cannot provide its own acceleration rockets don't count rockets carry fuel the fuel is burned this exerts a force on the rocket and so this can affect its motion but any projectile that has no way of exerting a force say by using chemical energy or some other form of energy to convert that into a into motion these will experience projectile motion so think about just how useful it can be to know that when you meet the conditions required for projectile motion it is extremely predictable that motion can be extremely well described now i'm not going to show you the punchline to the steel ball and can demonstration in here we're going to do that in class but you will see that if you set the situation up correctly as i said aiming with steel ball at the can having them release as shown in the experiment earlier at the same time that in fact the steel ball will hit the can now the caveat on all of this is of course there's this thing called the real world and in the real world things are more complex than what i've laid out here you had better account for things for instance like air resistance and also when you fire a projectile off the surface of the earth and it's no longer making contact with the surface of the earth and because we live on a planet that's rotating we go through one day night cycle every 24 hours that motion of the earth the ground moving and the air resistance this adds real complexity to the problem of course in real situations in artillery calculations in thinking about designing something like a baseball stadium at various altitudes or where air atmospheric conditions are slightly different than they would be say at sea level or in another park you need to take these things into account when you're thinking about how many home runs might get hit where the artillery shell might land and so forth but over short distances with appropriately aerodynamic projectiles at low altitudes with weak air resistance all of this works remarkably well and these conditions while they sound very limiting and and yes they do take you away from the most realistic calculations they are nonetheless good enough assumptions that you can make a lot of progress with this subject let's talk about the path of projectile motion i said earlier that the path is a parabola and in fact i'm going to reinforce that and show you that now so again the path of a projectile in this kind of motion with a constant acceleration in one direction no acceleration in the other it has this common shape the parabola now the details of the parabola its height its width these things can vary based on the initial conditions of the situation how fast you fire the projectile how strongly acceleration is and so forth but in general this shape will be preserved assuming you can neglect things like air resistance and so forth the equation for a parabola whose apex or as is shown over here on the right whose minimum point is located at some point in space and i've denoted that x apex and y apex and with a focus a distance f from the apex so the apex would be here and the the focus might be somewhere up here again review the definitions of parabolas that you might have learned in high school for instance to refresh your memory of all of this stuff the equation for a parabola with these conditions is as follows the y coordinate of the parabola is given by one over four times the focus distance times the quantity x minus x apex all squared plus y apex so this allows you to define a parabola that's offset from the origin and raised above the origin for instance with because it has this apex location that isn't zero comma zero but if we set the apex to be zero comma zero we recover a much simpler and perhaps more familiar equation for the parabola that you would have seen in high school y equals a constant times x squared and that constant is one over four times the focus this is another way of representing a parabola this equation over here is perfectly valid it's a slightly more general form of this equation with the focus and other pieces like the displacement from the origin and the x-axis and the height above the x-axis that the parabola starts etc that's that's shown over here on the right in this equation these are all equations for parabolas of various kinds generalized or otherwise what about motion well if we take the equations for x and y those coordinates for the steel ball at any time t after it's fired and if we rearrange them so that we can solve for t from one of them and plug that equation for time into the second equation we then find this eliminating time from these equations we only have y x x not y not v zero x v zero y and g and what do we see well we have the the height of the parabola above its initial starting point or below its apex is given by a term that is linear in x multiplied by the y component of velocity and that looks very similar to the b term over here in the parabola equation and another term that is quadratic in x and that looks very similar to the a term over here in the equation so we see in fact that the relationship between the y coordinate of the steel ball and the x coordinate of the steel ball at any time t after it's launched is given by a parabolic equation so this is just the equation for a parabola in a slightly expanded form over the generic equation specific to our projectile motion situation so again neglecting things like air resistance the rotation of the earth under the projectile etc the motion of a projectile under these conditions is completely predictable before we close out the lecture let's play around a little bit with projectile motion in a computer-based simulation a sort of game-like environment i have here a cannon that's capable of firing various projectiles and you can see that it's initially set to fire at an angle of 80 degrees above the horizontal there's a target that i'm supposed to hit by shooting projectiles out of the cannon i'm going to move the target over to about 20 meters from where the cannon is located so right about here so i can start by firing a projectile from the cannon i've selected pumpkins with a mass of five kilograms a diameter of just a little over a third of a meter i'm going to fire them from the cannon and watch as they execute projectile motion and see if i can hit the target so let's fire a pumpkin here goes a pumpkin up into the air and we see that it leaves a beautiful parabolic trajectory as simulated by the computer in the air the pumpkin smashes here on the ground we are far from our target we clearly missed the target but i can adjust the firing angle of this let me try firing a little lower maybe 56 degrees and fire well that was terrible this clearly overshot the target but you can see again we have this parabolic arc albeit wider the pumpkin flies further along the horizontal direction although it doesn't go as high in the vertical direction because of the firing angle of the pumpkin and the effects of gravity during the motion it's clear i need to split the difference and do something in between here so let's try something more like 65 degrees and fire the pumpkin one more time ah so close but again you can see that i've increased the firing angle relative to the ground of the pumpkin and so the pumpkin goes a little bit higher but not as far as it did before and you can begin to see these trade-offs in the motion of the pumpkin in projectile motion if i aim the pumpkin higher into the air when i fire it indeed it will go high up into the air but it won't go as far horizontally if i aim too low it'll go far horizontally but it won't go as high as that otherwise would and so hitting the target is a careful balancing act of firing up into the air and firing too low let's try one more firing angle let's try 68 degrees and see what happens so now i fire so close let's try one more time we'll try 70 degrees you can try this game on your own these simulators are available aha i got into the outer ring of the target not too bad these simulations are available from a group called vet vet.colorado.edu i'll make sure to put a link in the slides around the course website and we can probably improve our game a little bit more by maybe switching this to 72 degrees and firing let's see if we can get dead center on the target not too bad clearly the magic number here is going to be 71 and if i fire one more time indeed boom i'm in the sweet spot of the target so this will give you a little opportunity to play around with projectile motion without hurting anybody or yourself and you can try various projectiles you can also do things like turning on air resistance so as a last test of this let's try air resistance we're going to add resistance due to air now so the pumpkin is going to feel a drag force as it moves through the air i'm going to fire at the same angle 71 degrees let's watch what happens oh indeed we see that adding in a bit more realistic aspect that is resistance to motion from air greatly alters the trajectory of the pumpkin it still looks parabolic but it's not a perfect parabola anymore and in fact we fall short of the target because our motion is being resisted by the air through which the pumpkin is moving so feel free to play around with this this is an excellent way to get some intuition and a game like environment for physics and it will give you a more hands-on feel for projectile motion when you can neglect air resistance and think about trying to hit a target by balancing how much horizontal velocity and how much vertical velocity you present in the presence of gravity let's review the key ideas that we have been exposed to in this lecture on projectile motion we've seen that projectile motion is a special kind of motion that happens when motion in one dimension is unaffected by motion in any other dimension the motion of airborne objects near the surface of the earth is a really excellent example of projectile motion think back to the beginning of the video with the basketball player jumping into the air to dunk the ball or the pitch of a softball toward the batter or the throwing of a football to aim at some distant target make that shot as best you can these are all examples of projectile motion gravity gives us an outstanding laboratory near the surface of the earth to explore this kind of motion and we have seen that we can describe such motion using ideas that we have already developed regarding position and time and velocity and constant acceleration in cases of projectile motion where there is a constant acceleration along only one of the coordinate directions we have seen that the observed trajectory of motion in space will be a parabola we have seen how to make approximations for instance the removal of air resistance from consideration removal of the rotation of the earth underneath a projectile that's been launched from the surface of the earth and we've seen that for certain limiting cases and assumptions it's it's not a bad thing to do we do need to be cautious in the real world these forces can play major roles especially over large distances or with projectiles that are not very aerodynamic projectile motion is seen to be a beautiful kind of motion a predictable kind of motion one that we can not only describe with the mathematical tools that we have developed so far but whose outcomes we can predict quite reliably as long as we have made a valid set of assumptions