 Everybody welcome to Tutor Terrific today. We're going to look at our third lesson in chapter 3 of my physics course on two-dimensional kinematics This lesson is going to be a little bit heavy. We're going to derive the 2d kinematic equations This is an analogy to the big three equations from the previous unit, which was one-dimensional kinematics We're going to drive those today. We're going to use them today, and I'm going to derive another formula Related to a topic we learned in the last lesson called range. You're going to get a formula for that So let's get started So I just want to show you some pictures of objects undergoing projectile motion or an approximation projectile motion in real life It's in front of the Blasio in Las Vegas, the water show on the fountains. It's really neat to see And if you look real closely, you can see that all the water droplets before air starts to really take effect, air resistance You get to see that parabolic shape. You can see those half parabolas usually mostly in this picture for each of those water droplets In this next picture, we have a ping-pong player ping-pong championship and you can see four different snapshots of the Ping-pong ball we're taking and you can see that it's in the air and it's a partial parabola shape. I Told you guys in the first lesson I ever did about Volcanic activity and eruptions and the lava flows Notice the particles in the air are undergoing projectile motion. They're all parabolas shapes upside down parabolic shapes and that's because If you neglect air resistance, the only force acting on those is gravity but notice how we Seize the parabolic shape for the lava that is traveling down the side of the volcano. That is Definitely not projectile motion at that point because there's many more forces involved And here we got this funny gift of a car That this is in reverse this video of this car in some backyards and large backyard this old trashy car and We see even in reverse the the shape of the Car when it's in the air the shape of its trajectory is a projectile Even in reverse. I'd love to have a backyard like a doe with some junk car, but I Live in a place that doesn't have that kind of space Alright, let's get more serious here. So I'm gonna begin the process for driving the two-dimensional kinematic equations By making a particular important point. Let's look at this first tennis ball here Let's look at the components of its velocity. Okay, here they are and let's look at a tennis ball farther in time And it's sorry this one components of velocity Notice how we always analyze them separately. Okay, we we we looked and analyzed the horizontal and vertical components Separately you can derive the equations separately as well And we're gonna do that it makes it a lot simpler than looking at the entire vector for velocity For example and driving some equation for that in two dimensions. So we're gonna look at each dimension separately Let me show you an animation that makes it clear why we'd want to do that So on the left you have the green vector. That's the entire velocity vector and the red vector Stands for G the acceleration due to gravity that vector changes angle and magnitude at each instant in time On the law on the right side over here. We have the components of the green velocity vector So you could see the blue one for a horizontal as we already know is constant in magnitude and direction while the Vertical component of velocity the yellow vector is the one that changes magnitude But its direction is either up or down So if we were to separate those two components it'd be a lot easier to deal with then To look at the entire vector together because there's so much changing at once We'd have to have an equation for the angle with time and you have to have an equation for the magnitude with time That's more complicated. So that's how we're gonna do this So let's start with the horizontal component of velocity Notice how it's constant. That's the point. I'm trying to make here this vector is constant in magnitude and direction Because of that. We have a very simple equation for the horizontal component of velocity. It's just the Horizontal component of velocity as a function of time is just the initial value Vx not notice how I'm using two sets of Subscripts, I know I'm not the same size You'll see a nice version of this with nice math script later in the lecture but notice how I am now using a subscript x and a Subscript for initial value so that I know I'm talking about the horizontal components Because the horizontal component of velocity is constant the equations rather simple for the horizontal component of velocity Now, how do we get the horizontal component for position? We will need that equation, too We could start with the big three equation number two the long one that had time squared as the last term Just set the acceleration in that equation equal to zero and you'll get this X of t equals the initial value of x plus the initial value of the velocity x times t so that's the initial Horizontal component of velocity just like in this equation higher up. So this is What what we are going to do and we are going to definitely need to use a subscript to illustrate what we're component We're working with when we have component based variables and we often do so these are this is it for the Horizontal equations just these two and the first one's really not even an equation at all It's just a statement that says the horizontal component of velocity is constant All right now the vertical equations as you can see the vertical component of velocity is much more complicated This is going to take more than one slide for sure so They're definitely not as simple just recall that this should remind you of free fall from the previous chapter in the last lesson I did Lesson five in the previous unit. We talked about free fall So we're going to be thinking of free fall in chapter one to derive these equations Okay, there should be some similarities between the big three applied to free fall and the vertical components The equations for 2d kinematics for projectiles for sure So let's go back to the graphs to get started Here's a graph of the vertical position So the y position of a projectile with time that starts and ends at the same height So it's like you're throwing a ball in the air or something it takes some time to come back down So that's why it looks like a nice parabola. It takes some time before the initial and final positions are equal again This would be the equation for velocity notice how the slope of position is Positive and then it's zero and then it's negative notice how the values of velocity start out positive Then there's zero right underneath where the position is flat and then they're negative afterwards and they get more negative with time If we were to graph the acceleration, it would be a constant negative value because the slope of velocity is a constant negative value And so there's the acceleration curve right there We're going to start with the acceleration being constant. Okay, what's the exact value of that acceleration at the earth's surface? It's g Negative or positive in this case negative 9.8 meters per second squared So we're going to start with that it's constant the acceleration with time is a constant and it's equal to g Now it depends on whether you choose positive downwards or upwards as to the sine of g But g stands for positive or negative 9.8 meters per second square the acceleration cannot change and Projectile motion it does not change So what do we do to go backwards to get an equation for velocity? Well, we do what we did when we derived the big three We put a t near the original constant value for acceleration So it's g t and put a slap on a new initial value for the velocity So velocity y so this is the vertical component of velocity With respect to time is equal to the initial value of the vertical velocity plus g t So we have our first big equation from the vertical set now. We can go back again One level again and get the vertical position with time We would slap a t onto the constant from this equation. So it'd be v y not t This guy t will become t squared in this term and a one half will be placed in front again When I'm doing the understanding of what I'm doing and why you do it requires calculus Those of you are in calculus You know what I'm doing and then we'll slap on a new initial condition for the position and then we'll get this This looks very similar to number two of the big three equations from the last unit So we've got the vertical position with time equals the initial vertical position Plus the initial vertical velocity times time plus one half g t Square all right. We've derived these two These are two of the three vertical equations now There will be a time independent equation to round out the set of three Will we do it the same way as we did it for chapter two? Yes we'll start with the two equations we just derived and substitute out t in Both equations using algebra and you will receive we will achieve this equation Which is very similar to the time independent equation for one-dimensional kinematics Just notice how I have little subscripts v subscript y v subscript y not We're dealing specifically with the vertical motion. This does not apply to the horizontal motion Okay, so this is called the time independent vertical equation. So here they all are as a set look closely Horizontal motion you have those two equations like I derived and the first one's really just a statement of how the velocity in the x direction is constant notice that the subscripts look a lot nicer on these and I'm going to show this at the beginning of all the rest of the Lessons in this unit so that you can see them and I'm going to show them by all the problems we do So here are the two-dimensional kinematic equations of projectile motion I have one more equation to derive, but I'm going to save that for a little bit later in this lesson So g you notice how a has now been replaced with g in the vertical equations Stands for positive or negative 9.8 meters per second squared And we will be setting positive and negative directions in our problems and that will determine the sign of g All right, so we're going to do some example problems And we're going to do some example problems for the rest of these lessons But I'm going to start with two easy ones today. However, I want to lay some ground rules for you It is very important to realize that You're going to create your given and wanted tables like this says here But you're going to have two sets You're going to have a given and wanted for the x direction and you have a given and wanted for the y direction You have to have both and need to keep track of all your variables. There's many now that you need to keep track of Also, you will need to choose a positive and negative direction not just vertically or not just horizontally But both vertically and horizontally is left to right positive and it's down or up positive. You've got to do both Okay, now if there's no picture If the problem you get does not have a picture you need to draw that picture and it behoove of you to also draw the intended Trajectory for the projectile whether it's a full parabola or half parabola or a subsection of a parabola It's good to draw it and get some initial values for angle for example or the velocity vector Things like that and label your distances give it to you as well. So definitely draw a picture also Just a word of caution. I'm going to do the problems in the slides But I might skip some small level algebraic manipulation steps or derivation steps Because I trust you guys and you'll see some more summaries of those steps I just don't want to clutter the slides and so that's why I made that decision. Let's do it Okay, first problem. We're gonna drop something from the Leaning Tower of Pisa again 140 meters off the ground We're gonna drop it straight down from rest. We're just gonna let it go I ask you two things first with what speed will the object the ball hit the ground Okay, and part B asks how long will it be in the air before it hits the ground? Okay, I apologize for the grammar error there So here are our equations that we know We are dealing completely with vertical motion here. So the horizontal equations won't matter so much Okay, so the first thing says we have a picture of the situation and the problems picture already tells us to choose positive Downwards we're gonna create our set of given and wanted tables Here's what they should look like a given for the x direction a given for the y direction a Wanted for the x direction and a wanted for the y direction now if I set the in if I set positive downwards It makes sense to have my initial y position be zero at the top Okay, and the my final y position at the bottom to be the hundred and forty meters in the x direction I have no changes at all no velocity changes or position changes because I'm Falling straight down so all my x and vx values will be zero X and x not x final vx not and vx final will all be zero So here's what you have I've created both those sets V y not the initial velocity since I'm dropping it That means just letting it go from rest. That means the initial vertical velocity is zero. Okay, and Positive downwards means g. This is a vertical Term here vertical parameter g is positive 9.8 meters per second squared since we're going down as positive Now let's talk about what we don't know we do not know time We do not get any time information so we have no idea how long it takes and there's a tip to that It asks that in part B. Also. I don't know the final Velocity in the y direction are all fast. It's going right before it hits the ground two things I'm being asked for so those are the two wanted values now t is a little hard to place because t in The horizontal equations is the same t. That's in the vertical equations So you could put it in either list of wanted or just Centered between the two lists so that you know it's a part of both. I leave that completely up to you So in this situation, we know quite a few things. It's very simple and we're just asked for two things in Part a it says with what speed will it hit the ground? Well, that means we're finding v y f okay So we're asked for v y f in part a based on what we are given We need to figure out the road map Okay, we need to figure out the roadmap though. I'll allow us to find it meaning which equations do we use? Okay, and I talked about this in the last few lessons of unit two We've got to figure out the equations and then naturally manipulate them To get the variable we want knowing all of this information in the Y Okay, the X doesn't really matter because nothing happens, but in the y direction we have changed Knowing the initial final positions the initial velocity and G Which equation would I use to get v y f? Well, since I don't know time. I'm limited to the third vertical equation the time-independent one Okay, that's our roadmap Now we need to manipulate this in order to get v y final, but v y final is already alone It's already isolated. So I'm just going to plug in some zeros I know that the initial position y not is zero and I know that the initial velocity in the vertical direction is also Zero so that simplifies the equation to just v y f squared equals 2g y final Then I need to square root both sides and when I square root both sides since I'm just plugging in some values over here This is just a number. I need to do a plus or minus when I square root. However, I'm not expecting a negative velocity Because I'm falling and I said positive is downward. So I'll take the positive Solution to this and then I'm going to plug in chug I'm going to plug in 9.8 positive meters per second squared for g and 140 meters for y f And I get when I square root that 52 meters per second. It's moving quite quickly As it right before it hits the ground from that height So part a done. That's how what speed it will hit the ground with 52 meters per second How long will it be in the air before it hits the ground now? I'm looking for time, but what I want to do before I find time since I know v y f I'm going to move this quantity to the known given y quantities So I could see that I now have that information to use T is the only thing I need. It's the only thing I need. So what's the easiest way to get it? We cannot use this horizontal equation because we'll just get zero because every term in here is zero There's no change in the horizontal direction So we go over to the vertical motion graphs and we then the equations and we look for the one the simplest one That allows us to find t that would be the first one. I have v y now I have v y not and I have g so t's the only unknown So we're definitely going to use that first one to solve for this Okay, I can already plug in that v y not at zero So it makes it even simpler. All I'm going to end up doing is multiplying g and t together But I need to divide by g because I'm solving for t Excuse me, and so we're going to divide by g going to plug in 52 meters per second for the final y velocity Divide that by 9.8 meters per second squared for g and you get about 5.3 seconds Which seems a little on the long side, but 140 meters is quite high and so That would take some decent time to take even more time if we allowed air resistance to create drag Okay, so there's a two-part problem for you real basic because we really don't have any motion in the x direction That's the last time you're gonna have a problem. That's simple Let's talk about a driving off the cliff example like we did Visualizing wise in the last lesson. So we've got a movie stunt driver. He's on a motorcycle He speeds horizontally up a 50 meter cliff, okay? So he's driving he's traveling directly horizontal as soon as he leaves the cliff's edge. It's 50 meters high How fast does he have to leave the top of the cliff in order to hit? A ground level spot 90 meters away from the base of the cliff That's where the cameras are and so he wants to hit that It's not drivers have to do this all time that to calculate the velocity that to leave the cliff or the jump with To hit that specific target Okay, so we have this 50 meters and this 90 meters and we've got some other information too, but If you don't understand what's going on or you haven't already done so, please draw a picture I'm not gonna draw one here because I don't have room on this slide However consider the cannon Ball being launched horizontally from the cliff that was an example in the last lesson when we were drawing our Vertical and horizontal component velocity vectors. So that's really what's going on here It's half of a parabola trajectory And the 50 meter top cliff and then 90 meter distance Now you have that you're gonna create your setup given in wanted tables before you do you need to set a Positive and negative direction for vertical and horizontal motion I'm gonna say based on the fact that I like things traveling from left to right I'm going to have positive be upwards and I'm going to have Positive be to the right you'll see why I chose positive upwards based on the solution. That's what was done Normally we might choose positive downwards because the bikes never traveling up But I'm going to show you that it's fine to choose Either direction as positive you just got to set your signs correct for the given and wanted information Now let's talk about that given X. I know that my Final X position is 90 meters away from my initial X position So what I would do is I'd set my next my initial to zero meters and my final to 90 meters from the base of the cliff also I Do not know my initial Vx My Vx not I don't know it and so that will not go in the given set So it's only those two position pieces of information that go in the given X set Now given why I know I'm going to fall 50 meters Okay, I'm going to start 50 meters up and I'm going to go down to zero meters at the end. Okay What I'm going to do to Match the solution that I got online is I'm going to set the top of the cliff as zero and If that's positive upwards and I fall 50 meters the final Y position will be negative 50 meters You'll see why I had to do that also I know always know G guys do not put G in the wanted lists G is always positive or negative 9.8 meters per second squared and since I chose positive upwards G will be negative 9.8 meters per second squared and Since I'm speeding horizontally off the cliff. I know that my initial vertical velocity is zero All right now, let's go to wanted X and Y Well, I'm asked for Vx not how do I know because it says how fast was the motorcycle leave the cliff top? It's the stunt driver speeds horizontally off the cliff So I know the entire question is asking for Vx not because be why not is already known Okay, also the final velocity in the X direction, which will be the same as Vx not is technically not known as well. I Also, don't know the final Y velocity The vertical velocity that this bike has and stunt driver has right before it hits the target I also don't know the time. I'm not given any information about time once again so Look at the roadmap How can I get Vx not? Well, I don't see Vx not in any of the vertical equations. I see it in one of the horizontal equations Well, two of them one of them. I can't really use because it's just a statement the second one I can I need to know the initial and final horizontal positions, which I do But I don't know time. Oh There are two unknowns in this equation. I can't start with it Looks like in order to get Vx now I have to find the time some other way and that requires you to go over to vertical land and Use one of the vertical equations to find time. Okay, so which equation allows us to find the time? Given things like the initial and final position initial velocity well Let me look closely at the second one the second one Has that information. I can't use the first one or the last one For different reasons the first one because I don't have Vy final the last one because it doesn't have t in it So why would I bother? It's the second one. All right now I'm going to set that here and I'm going to plug in some zero values I know the initial position is zero so that can be a zero I know the initial velocity is zero in the y direction. So that's a zero as well so this situation got quite simple quite simple and G was negative 9.8. So it reflects out here and I see that this just becomes y equals negative one half g t squared very simple way to get t just going to plug and chug The g and the y final and solve for t Okay, and so Doing that we're going to multiply both sides by two and then we're going to divide both sides by g and with the negative and So if I check the way I've set this up y is negative 50 meters So the negatives cancel out. You don't get an imaginary value for t. You get a real value And it's 3.19 seconds Okay, now I have the t From the vertical equation number two. I'm going to take that t value over to the horizontal motion I'm going to plug it in along with x and x not and then that's going to allow me to get my final horizontal Initial horizontal velocity value that I need The initial position is zero. So that term was also emitted So it comes down to simply dividing both sides by t Plugging in the final x 90 meters and the t value I just solved for it 3.19 seconds This gives you three sig figs for your answer 28.2 meters per second Bike has to be traveling at that specific speed to land 90 meters away if we ignore any resistance due to error anything like that So this is an example of a problem in which you saw the need to use more than one equation One equation just to get you a value that you needed in order to get the value you actually Wanted you were never allowed to use equation that has more than one unknown to solve for something You have to have everything known except one thing in any equation to be able to solve for it All right, so enough problems. We're just going to derive quickly a formula for range Range is that difference between the final and initial horizontal positions? Okay now There's a specific Parameters that need to be taken into account in order to do this. You can't always use this formula But let's see where it comes from It comes from the equation number two in the vertical set and it comes from the only real useful equation in the horizontal set Here's what you have to do To in order to derive in a formula for range you have to set the initial and final y values to zero They have to be the same you set the initial x value to zero as well When you do this you can work with these equations use a little bit of trig There's a little bit of trig in there. I'm not going to show you everything I'm going to show you the final result though The difference between x and x not ends up being the total velocity Not any component the total initial velocity vectors magnitude squared times sine of twice the initial launch angle Divided by g which is 9.8 meters per second squared You are only allowed to use this for what is called level horizontal range level range the initial and final vertical positions Must be the same or this can't be derived. So this is our what we call range formula Okay, so I'm adding this to the set of all the equations we need. This is it for the unit. This is it We're just going to practice using them. We're going to get better using them some more complicated problems we're going to see what they can do and That's all we have. So I've finished showing you and introducing new material to you This is your two-dimensional kinematics equations for projectile motion All right guys next lesson. We're going to continue to do some problems But thanks for watching this video for now. This is Falconator signing out