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2. 2d Kinematics 6 Angled Projectile Motion High School And Ap Physics 7: right move. So here let's talk about angled project emotion. And when we talk about angle project emotion, we're simply talking about any projectile that it's fired at some angle. Hence angled project emotion. Remember in the previous examples we gave me draw that real quick, we gave horizontal project emotion. That meant the initial velocity is completely horizontal, Right? Angled is simply projectile where its initial velocity is at some sort of angle. So the classic style would be like a soccer ball kicked off of the ground at some angle to the horizontal, which is the ground. But we could have a few other similar type of examples which are really no different in terms of the physics, but might look different. So maybe we have that cliff example, but now might be the object is leaping off at some sort of angle. So it goes up first and then comes down and we could have the leaving from the ground level scenario where is launched to hit some structure, maybe a window or a platform above ground. And it's similar, of course, to the this one over here, and this one over here. So this is angle project a motion. The physics is all still the same. It's an object travelling in two dimensions solely under the influence of gravity. However, because it's a little bit more specific, we can give Ah, a little bit more tips and some concept that goes with it. Ah, the angled ones There may be typically considered to be a little harder, a little bit more difficult, because you're going to have to take into consideration. There is an initial vertical velocity, whereas in the other problems there wasn't. So let me go ahead and draw out, um, the projectile, the classic angle projectile motion style where I've got some object we could think of it as a soccer ball kicked with some angle to the horizontal, and it's going to project through the air, and in this example, it's gonna land at the same height in which it left. And the reason why I'm giving this specifically is because it's easy to kind of explain how the projectile changes on its way up to its way on its way down. So let's look in this scenario here where over here, this is the total time it took for the projectile to travel across the way. And here t is zero. The first concept I want to give you is dealing with the peak of the projectile when that object, which I will draw in a darker green than that background green when that object that its highest point. So when it's up here, there's a few things that we need to understand. One. It's vertical velocity is zero to the time it takes to get to a tires point will be equal to the time it takes the fall from its highest point back to the same ground level. So long is that ground level is the same, of course. Okay, so here we're gonna call it t 1/2 which is half of the total time. So if I knew it took four seconds to travel all the across the way, then it only took two seconds to get to its highest point. The third thing, understand? And this is, ah, pretty paramount that this horizontal velocity here is the same as the original horizontal velocity ball had in the beginning. So let's look at the beginning. Vector. Now let's take a look at this guy here. This vector here, which I'll now go over in a darker red is our initial velocity vector. That velocity vector should get broken up into its component vectors the X and V i Y. Simply because in the very beginning, this ball being kicked in an angle has both that X velocity in that y velocity. And remember your X and Y values rather its velocity, acceleration of displacement or any other motion variable your accent y values never mixed together. So since B we have both x and Y in the beginning, we will have to analyze how they both change separately from each other. So this V X ray here is the same as this Vieques up here. Why is that? Well, again, as you can recall from the horizontal motion projectile motion video, your ex component will never change because there's simply no acceleration in the X dimension. There's no force acting against it now. Of course, if there was something like air resistance, we would have to at least concert, you know, conceptually understand how that affects it. But in reality, we're gonna be dealing with no change in my ex velocity. So wherever I draw this projectile Avila have the scene. X velocity vector everywhere is this y velocity vector, which will decrease on its way up in an increase on its way down because gravity only affects the wide values. Keep that in mind, right, That acceleration due to gravity. All right, Cool. Um, before I get into talking about how the individual vectors changed, wrote its path, let's just quickly talk about what does happen if I have a resistance. If I include air resistance, as you recall from horizontal project on motion, the object will fall or finish a point in time closer to where it would normally have. So our range is decreased because there's that air resistance acting against in. As the project moves to the right, it will be slowed down visa via resistance. How will it affect the maximum height? Yeah, it's gonna also decrease. Our maximum height will likely be smaller than if we didn't have a resistance. Because as a project, I was travelling up through the air. Air is resisting that path as well, so both its maximum height and its range will change. Got it. I've got a lot of stuff going on here. I'm going to raise this and kind of give myself a fresh example our fresh projectile path so I can kind of muck it up again by trying on it again. So I'm gonna give myself my ball, and it is going to travel through the air in a nice projectile path, and it's gonna win some distance away from it will call that X distance the range. And I just want to draw this ball at multiple spots. So I'm gonna go back in time to do the highest point. And I'm not just say again here and here. And ultimately, I just want to write down on my vectors at those what is five positions? So the very beginning, we did have both the X and my initial. Why now? I don't need to see the I X cause that'll never change. But I do need to see v i. Why? Because that will change. So this is what I have in the very beginning. Okay. On my next moment in time, okay, anywhere but at the spot that I've drawn my ex value won't change. As I have said multiple times, X velocity will not change unless, of course, there's a force acting against it. How does the why component change while we're not at our highest point? So it's still traveling upward. But will it be traveling upward with the same initial life speed that it had in beginning? No, this is because Gravity is acting down at this stage. It's acting down the hallway, right? So it's gonna decrease its why Velocity? So I'm gonna say V two y is less than V one way. Now we're at my highest point. Well, the object is still traveling to the right else. It would have just fallen straight down, right? So V X is still here. And it's the same values that was here and here and everywhere, and it will be the same everywhere on the way down. What about my y velocity here? What's view y three? Well, we already said it or V three. Why I said that before is indeed still zero, or is indeed zero where at its highest points, no longer troubling up. And at that instant it's not flowing down either. And of course, the next sentence than in time. It will start to fall, and then we get right here and I want to draw this value. And I tried to make this in line with this ball doesn't look like it's exactly in line, but just pretend it is. Ah, it's meant to be the same position in its path on that's done for a reason. And that's because I want to point out the symmetry to project on motion that same point time after my highest peak, to the same point in time prior to my highest peak, my velocity vector magnitude will be the same. The four y magnitude equals V two y, and this is because gravity pulled it down the same way gravity slowed it down on its way up. So on its way down, it gave its speed the same value that it took away when it was on its way up. The direction, of course, is different rate V four wise pointed downward, but the magnitude of the two are the same cool. And at the very end, we still have V X right here. And how does V five white compare? What is that gonna be equal to, at least in terms of magnitude? Yeah, it's give me equal to be one way these two vectors are going to the same magnitude. But, of course, opposite direction. Got it. No early stuff. The last one I want to point out is kind of a trap question. And I see this happen all the time more often in the high school level. Physics problems and you'll still see in the AP style questions. And it's just simply say, at the highest point in its path. What is the acceleration of the ball? And so many students want to say here, a zero. No, that's not true at all. The acceleration here is not zero. If the acceleration Alitalia's point, was zero, then the ball would never come back down. It would cease to change its motion. It would literally just travel to the right forever. I would be crazy if when you kick a ball at its highest point, the acceleration acting on it disappears. Because then the ball would literally just travel forward forever. We'd have the history, the world's history of objects flying around horizontally above us because they never would have come back down. That's simply not what happens. The acceleration, its highest point, is still little G. It's still 9.81 meters per second squared and it's down. So we're gonna say it's negative and that's the case everywhere. So if I were to draw an acceleration vector everywhere, it would be the same size and it would be, Ah, the same everywhere. Keep that in mind. The acceleration vector is G everywhere in its path, and that is, in general, the basis behind Project on Motion. Now I caution this Next thing I want to write down is extremely, um, it's really important. Understand that it's on Lee for specific search circumstances. So everything I just explained to you here's legit for all angled problems. But specifically when my change in why displacement is zero i e. We start and finish the same location height wise. Of course, it's traveled further to the right. If my change in displacement in the UAE dimension zero, we get to use a specific formula to allow us to calculate the range and it's on Lee, true for this set up, so I wouldn't necessarily encourage you. Just memorize it for all of project I motion just memorize it for this particular situation , but it will be helpful to solve or to use, and that's the range formula. The range formless says that R is equal to V. Squad V Square. Signed to theater all over G. This V is indeed the vector at the angle. So my original velocity vector not either of the components. Fada is the original angle to the horizontal and of course, GSG. Now the sign to think that just simply means take fate and double it before you sign it. So if my ankle was 30 degrees, I would do the sign of 60 million was 45 degrees would do the sign of 45. OK, g is baby G 9.81 meters per second squared and V is whatever velocity vector gives you. This is Onley to determine range, nothing else. It won't help you determine height, maximum height or anything like that's just for range, But it can be helpful instead of having to go through a relatively tedious path to figure out the range itself. Okay, this is the basics behind angled project emotion. I'm gonna actually cut the video now and have a couple example problems in the next video. I think that it's gonna take a long time or a long enough time. I just want to break it up. Hope this help. Thank you.
3. 2d Kinematics 7 Angled Projectile Motion Example High School And Ap Physics 8: Hey, Moosa, here this video, I'm going to go over a project emotion problem. That's an angle. Okay, so I'm not gonna write out the problem. Just get a kind of say it out loud drunk right down my variable. Something to say. I have a ball that has an initial velocity of 10 meters per second. Kicked at an angle of 33 degrees to the horizontal. I want to know a bunch of stuff. I want to know what is the maximum height it gets to. So instead of me saying, d Delta, do you want actually gonna come back to explaining what I mean? There s So what is its maximum? I want to know what is its total range. And within that I also want to know the total time of flights. These are my This is my set up, and I'm just gonna sketch it real quick. So I have this ball kicked in it. Lance, this is 33 degrees. I want to know this Delta D Y. Yes. I'm gonna come back to that Technically dealt the d Y zero, right? So really, I should be writing h for Heidi. I'm just gonna stick with Delta do y? Because I know what I'm talking about here. I want to know my arrange Don't the DX and I want to know the total time it's in the air now. In the very beginning, this ball was kicked at some angle. So I've got this is view I, er v one That's the 10 meters per second. In order to solve an eighties problems, we're gonna have to find our X and y values. So I think it's really important to create an X Y table. The gross majority of physics teachers out there explain. Tell students to do this. I am the same way. I genuinely think it's the right path. So typically, other than reading it, drying a sketch is starting to make a table. Your real first, Um, step is the Find your components for Exeter for the the this vector here at the 33 degree angle. So before even write down, um, Minoans list in a table format, I'm going to find and understood do this off to the side. So I have more work space on the defined V X and V one. Why? So I need to find the X in view one way. Okay. And to do that, I got a look at this triangle and I got to think about my trigonometry. And I got to remember that so could tow a stuff. So if I may be, if I write this down appear, it'll help me remember it. So, uh, to, uh and let's look for my x component first. Something to say. What is my ex component of this velocity vector? I'm looking at this right triangle. I see my velocity vectors by Pont news. This X component is adjacent to my Anglo Fada. So I'm going to use the one that uses a Jason iPod news. I'm gonna say cover sign, and I'm gonna jump to the rearranged version. I'm gonna say V X is equal to V. Hussein Stada, which is gonna be equal to 10 meters per second times. The co sign of angle 33 which is it's tosses in my calculator here. Ah, eight point around that to 8.4 meters per second. Nearly got. Maybe y we'll be wise opposite of my angle. Fator. I have my high partners. I'm looking at sign so v wise get equal Toe V Sign theater or 10 Sign 33. Let's put that in the calculator, and that's gonna be 5.45 meters per second. Awesome. Now, now that I have my components, I think it's really important to toss them into the appropriate X Y spots. The way I don't forget that that's what I'm dealing with. And I'm gonna go ahead and finish everything else I know on my knowns list. So I'm gonna put V X right here, and I'm gonna put V y right next to it. Reminder Pay attention to sign. So if they're both positive, write them down is positively on this case view wise up to the exit to the right. Also, I have two subscript my v y more specifically to be one. Why? Because that will change. And I don't even do that to Vieques. Let's go ahead and write down everything else we know and we look at our noses, we say, Well, we don't really know much. I mean, I could put Fada in there, but that's not really a variable. That's just something that set in the very beginning, so I don't need to worry about rating that down. So does that mean I know nothing? Probably not. I know my acceleration, my acceleration. The ax is zero. The object is not increasing or decreasing speed left or right. And I know my acceleration in the Why is G and I know G is directed downward on this object . So it's giving negative 9.81 meters per second squared that negatives important. Do not forget that negative. I don't know my range. So that's Delta DX. I don't know the time the object is in the air. I don't know my height and I don't know. Also the time it's in the air, right? That's on both sides. See, say okay, it's a tricky problem. How do I solve this problem? Because, remember, you need three of every variable to find every other variable, so that's gonna make it difficult, cause we only have two over here and we have to over here. So it must be something else that I need to figure out. Let's think about it. What is the motion variable that I haven't accounted for yet? If that's gonna be my second or final velocity, I have not accounted for V, F y or V F X. But VFX doesn't change right chili v F Y. I have got a look at, and there's one of two ways that I can teach you to do. This is one of two ways in which all of physics teachers teach. There. You get either good to say, one of two things they're going to say. Find this information at the highest point i e. What's my Y velocity here? Or recognize that my final Y velocity must equal my initial y velocity, but opposite direction so would be negative. I'm also noticing here that I never plugged in that negative symbol, and I probably should turn out negative symbol that my units for velocity they should have done that. I'm gonna go ahead and go with the traditional route. I'm going to say, Let's find the time and in the distance at the maximum height. So I'm gonna go ahead and say V two y zero. But I could have just as well said V. F. Y is negative 5.45 meters per second, and I also know the X here right is 8.4 meters per second. Um, I'm gonna go ahead and use the V two I If you want to go ahead on your own and try using V f Y, you're gonna get the same stuff and the reason why I'm gonna do the view to Y equals zero says they know any maximum height anyhow, and that's at that position. So hopefully you process what I just said because it's gonna be important. You're gonna be using this a lot. I'm going to say V two y zero. A lot of times I tell my students, even if it's just for their own good, a little make a little note. Just said that that's that Max each. It's not at the very end, that's at the maximum height. I think it just helps. That way you don't forget where that position is. So now what I want to do is I want to look at my variables and I want to think Okay, can I find D or T? Yeah, I absolutely can. I think I'm gonna find time first, because I know I'm gonna need that over here anyhow and then I'll decide what I want to do after, so I'm going to say what color will do this and I haven't done orange yet. I get really trained myself to not bounce around color so much. I'll figure that out eventually, right? I'm gonna go ahead and find time and a fine time. I'm going to say that because I know my initial my final My acceleration on this a V two y equals V one y plus a t And I could write GT if I want him. Whatever time is what I'm looking for. I know v two y zero I'm gonna subtract my V one y over and then I'm gonna divide a over means time all equal Negative, My initial I velocity divided by acceleration. I can't forget that Negative. It's going and figure that out. Now it looks like I'm getting 0.5 repeating. So I'm just gonna say 0.6 seconds or I'll say 0.56 So is that the total time is at the time it took to get back to the ground. No, it's not. This is the time it took my initial I velocity to get to zero under the influence of gravity. That's way up here. But if you remember the whole symmetry of angled projectiles, especially the ones where they start and finish the same height. The time it takes to get to the highest point is half of my total time. So I should be more specific. And instead of saying this is total time, this is half time. Total time is give equal this doubled. And since I saved my calculator work, I'm just gonna multiply that times two to being 1.11 seconds. That's my total time again. Here was the 0.56 seconds that times to the total time it took to get back to the ground. I can write that in both columns. Remember, time is scaler. The time it takes to go left and right is the same time it takes to go up and down because it's the same object. It's not like separating itself in traveling to Dimension separately. That's weird. Okay, now that I know a time, I could do one of two things I can use that time over here to find height. I could find height without time. And I can use the time over here to find X. I'm gonna go ahead and find my height first and some gentle with my wife values in the first place. Miles will keep doing with my wife values over here and then all the race, my work and duty X stuff, right? I think that's fine. If you don't like that, whatever. Yell at me. Okay, let's go ahead and do this one and a kind of a deeper purple. And I'm going to say that the displacement I don't know, I could do VF squared equals V I squared plus to a d Ah, I don't really like that too much because the v f n v i and there with signs, people speed it up, I'm gonna just go ahead and say Delta d Y is equal to D I Y t plus 1/2 80 squared. I know all of these variables. It's already set up in the D version for me. Might as well use it. So I'm going to say 5.45 times Now, this is my maximum height Maximum heights here. Am I going to use total time or am I going to use the halftime? Yeah, I'm going to use the halftime because it's at my Maxim White. If I use my total time here, I'm going to probably get a value of zero because there's no height at the very end. All right, somebody's 0.56 or 0.5555555 all way down, right plus 1/2. I'm gonna include that negative sign. If I don't include the negative, then it's basically telling you that gravity is increasing its effect to travel up and not decreasing its effect. It's terrible that times my time squared. All right, let's toss this on my calculator and see what we get. So 5.45 times 0.55 Now I need to say, minus this whole term, because that distributive minus 0.5 times 9.81 times and I like the ties my square stuff in parentheses just because I'm nervous that I'm scaring something else up. So I have to parentheses open here. I'm closing both practice. It calculated work. There's definitely more efficient ways of doing it in this. I expanded out. I go into the rule. Extra parentheses can never hurt. Okay, this is a maximum height of 1.4 around that toe. 1.46 meters. Awesome. You know what? I think I have enough room to do my final range down here, and so that's what I'm gonna do. I'm going to find my ex velocity or ex Velocity X displacement here at the end. So what is this Total range? I can only use X values. Do not combine your X and y's so we're gonna ignore this column completely. I know I have three, so I can do it. I'm gonna go ahead and say average velocity is equal to my change in distance over my time . I can do that because the velocity is indeed not changing. And I know that because my acceleration zero So my ex displacement is good. Be equal to my ex velocity times time, which is the 8.4 times. And, um, I can use halftime or total time. I had to use total time because that's the time it takes to get forward that distance. That's what I'm looking for. Someone say 1.11 This ends up equaling. Might as well costed in my calculator 8.4 times 1.11 were to go out and say nine point. I'm gonna say 9.3 meters. Awesome. That completes this problem. I think it's pretty good. This is, I think, the most difficult style problem you got to get at the high school physics level at the AP physics level, it kicks up a notch. So the next video I'm going to dedicate to Justin AP style problem. This will be it for this prom. Hope you like it. Thank you.
4. 2d Kinematics 8 Angled Projectile Motion Example 2 Ap Physics 9: Yo Moussa here, we're gonna do a project on motion problem. That's probably the more advanced problem that you're going to get at the college level. I could think of a few that might be a little more challenging than this, but this is right up there and you won't. You probably won't find this kind of problem on ah state exam for high school level. But you know, teachers might might drop. This is I don't think it's like, incredibly difficult and really, it's like, ah, horizontal problem. But we're starting off at an angle instead. So we're gonna be up on a cliff in this case some six meters high, and we're gonna have this object is going to leap or launch off. Maybe it's a ball rolling off a ramp or something. It's gonna get launched off at a velocity initial velocity of seven meters per second. But that velocity is 50 degrees to the horizontal, so it's not a horizontal problem. It's an angled problem and my question, and the only question is, where will the Obst land away? How far away from the base of the cliff with the object land Delta dx, and I'm going to do an X y table. I'm gonna have that whole the whole shebang. Um, the reason why this is more difficult is because we don't have that beautiful symmetry where halftime is equal. The total are is equal to ah, half of total time are Shelagh say it differently. The height, the time to get to the max height won't be half of the total time. That's because we're falling a great distance further before we get ahead of myself and walk you through it. Let me tell you what not to do. I see students all the time find the time it takes to get to here. And so that is a traditional, symmetrical angled projectile problem, right? Similar to the last video. Once you find the time it takes to get there, they then find the time it takes to go from here to here. But there's all sorts of murkiness involved in that. We've got to figure out how much um, you know, you have to encounter downward velocity here. You have to figure or figure out this. It just gets weird, like kids. Try to figure this range out and then find this range and at the two together know that you're going down the wrong path there. That's not what we're looking at. That's not I don't believe the right way doing it. You can do it that way as long Was it really disciplined and take, you know, write down things properly. What I think you should do is, well, I'll just show you. So first and foremost, figure out your X and Y velocity vectors in the very beginning. We know this guy has X and why, in the very beginning, its at some angle song ahead Go ahead and say V X is equal to V coasts data. So it's gonna be equal to seven coasts 50. And I'm gonna write out v Y woman down there to save time. V wise would be V signed feta, which is gonna be seven. Sign 50. Get comfortable identifying your trig stuff right from the start just by looking at it. The majority of Times X is co sign a y is signed, but not every time when we're in this quadrant one, it is fine. But when we have a velocity vector downward, it won't always equal that way. Okay, so just think it through. A Jason is X in this case, so I'm gonna use co sign all. We don't want someone to go ahead and say seven coast 50. Also helpful calculator reminder. Make sure you're in degree mode going around this the 4.5 meters per second and then I'll do seven. Sign seven Signed 50 and that's going to be 5.36 meters per second. They're gonna be similar. It's good to check this stuff through in your head. 45 degrees there, equal to each other. This is at 50 degrees, so it's not much bigger than that. At 50 degrees, we should have a slightly larger wide and we do it x at 30 let's say 40 degrees, which you have a slightly larger X than we do Why? And so this is 50 degrees. It checks out good. I'm gonna go ahead and plug this stuff in my X Y table on the same I X velocity in the very beginning is 4.5 meters per second. My wife velocity is 5.36 meters for a second. I know that my a acceleration zero he don't forget that I know that my y accelerations acceleration due to gravity. So negative 9.81 meters per second squared. Don't forget that I'm looking for d X. I'm probably gonna need to find time to do that. That's usually the case. I know Delta D Y. In this example, we were told that my total displacement. Now, here's another thing Mr is trying to do. They think that delta D y means the greatest point from the highest points they try to find the highest point first. And then add was No, no, no. Delta d y is the total change in displacement from the very beginning in time to the end. We don't care about it going up or down. Well, all we care is that it started six meters above the ground and have finished at ground level. So we have a total change in displacement of Well, I'm not gonna write down positive six or negative six. Did I go up six meters from beginning of the end, or did I go down six meters? I went down. This is negative six meters and again, I know it actually probably traveled a greater total distance in the UAE dimension. Ah, but we don't care about that. We care about the total change in position. Also, I caution you. If there was a platform here, let's say I had a platform. Don't write this down 70 race This let's have a platform is two meters high, right? That change in displacement is no longer negative. Six meters. It's negative four meters because it starts a positive six finishes at positive two final minus initial means we have a change in four. So I'm giving you a basic version of a difficult problem. But it can get more difficult. Sweet Ah, and also time. What is time and we don't know my final. Why velocity down here either? And I could have asked that. But for the sake of time, I'm not gonna. But I didn't want to just conceptually explain this at the very bottom right before hitting our target, we still have that X velocity of 4.5. But we now have a pretty large down velocity. This y velocity or V final is give a significantly larger than the magnitude of my mission . Y you know, back in the beginning, when we are in Sim. It symmetry my wife velocity magnitude will equal my initial wife lost city meant. But now that I'm falling further than I went than I started, I'm gonna have an opportunity to increase so much more speed than this whole extra portion s ovf wise. Definitely be larger in magnitude than V i y. I'm not gonna go and figure that out. But we could and then also again, if you want for the sake of heads up, not only could I find my V f y, but I could then find my overall velocity at the end. What is this V here? Let's not give you seven meters per second to be much larger than that. We've fallen quite a bit on and we could even figure out the angle below the horizontal. And for that once I knew the y and V X, I would just do pie fag and an inverse trig tan specifically to find data again. I'm not gonna do that for the sake of time. Certainly not improbable. So how how my go to figure out on? Of course, I don't really need time. I need DX, but how much you to figure out. DX will understand. I need three variables. Antley up too. So I have to find time typically to find time. I look at my why I do have three so I can find time first. So that's what gonna do. How do I find time? And a lot of students want to avoid this. Ah, but you're not gonna be able to. That's why I picked this kind of problem we're gonna do. Ah, the long equation. We write it out there to say Delta D Y is V I Y t plus 1/2 a t squared because you can't think of another equation. I promise you that will allow you to go around this unless you really break it up into many , many steps. But I don't I really don't think that's the rate path. So how do I find time? So I'm gonna say, let's isolate t. So now you start to do it and you go to get t by itself and you realize, Wait, I have an exponents in a non exponents. I can't. I can't combine these two t and T square. Don't go together. How am I gonna isolate t by itself? This is tough, isn't it? Let me put it in a format that might make it so you can see when I'm trying to get to I'm gonna have it as 1/2 a T squared. Let's get a kind of re rate this side over here. Plus I ve I y t Now I'm going to subtract this delta D Y over some negative delta D Y. I'm gonna sit this equal to zero. I want to look at that equation. Tell me if it looks familiar at all. It's okay if it doesn't because this is one of those first times where you're taking this mathlete this lead from math to physics and want to put a math hat on and tell me what you think this looks like. Let me write it down. Appear where I'm trying to go. We're gonna write it up here to say this. Do you recognize that a X squared plus BX plus C equals zero? Yeah, that's a quadratic. We are dealing with a quadratic you. We have two unknowns. One woman exponents. That's a quadratic. Now, if this ended up being a perfect square, I could dio and uncalled Reddick it. I'm not a math teacher. I don't know what term they use now factoring whatever. I'm not gonna do that. I'm going to go down the path that students like to avoid, but it's really not that hard. And I'm gonna use the quadratic formula. But before I do that, I want to connect this to this. So if I look at it, my A term is 1/2 a So my ex turn is t squared, right? My be term is V I y in my C term here. Is that change in displacement? Negative. The change in displacement. Sweet. Let me plug in my variable. So it looks a little clear. Somebody say 1/2 of a And for the sake of this video, I think I'm gonna treat a as now. I won't. I won't. I'll keep it Is 19 when I was gonna do it as 10. But might as well go the real way. So I'm gonna say 1/2. It's a 0.5 of 9.81 So a is 4.9 something to see. 4.9 t squared plus 5.36 t minus minus six. So it's plus six equals zero and my t I'm gonna put a little hook on it so I don't mix it up with the positive. Okay, hopefully that helped again. Check about Delta was negative, but the D was negative. Some subtracting, negative, making a positive students skipped that a lot on accident. It's definitely important. That means if I write this down off to the side, which I think probably makes sense, right, I'm going to say my A term is 4.9. My be turn is 5.36 my c term of six. Now those of you that have been completed trig yet or in Trig and I am a god of the quadratic formula, I will tell you this. This is the most difficult trick you'll do in the physics, the physics level. Unless it's a couch based course. So don't let this overwhelm you. Try to maybe even learn how to use a quad form that rate from this video. It's really not that bad. Um, you will get to the wise and house. And where's the quad form that comes from in your math class? I'm not gonna go do that. I'm not a nap teacher was gonna show you how to use it. So let's write down the quad formula. We're gonna write down X equals negative B plus or minus square root of B squared. Plus four A. C. This entire term gets the vital by two A. To I'm sure you've learned this before. This is in your math class. Who? Wait a second. Harris isn't. Plus, this is minus. Um, and sorry about that. Let's repeat this X equals negative B plus or minus square root of B squared minus four a c all over two a. Don't do what? I didn't make that. Plus, that would throw everything off. That be problem. Now why do we use this? We use is the soft for X. When we have a quadratic you can't solve for X linearly. This way. So we have to use this car quite formula get Identify your B A n c. Terms which I've done over here, which is just the terms in front of my unknowns. Plug it in exactly at face value. This plus or minus means we're gonna get to answers. You're always gonna get to answers with the quality for me. I'm gonna tell you this right now, in physics, we're gonna throughout the negative money to keep the positive one simply because in this case, for looking at time and time isn't gonna be negative. Okay, so this is the equation or use. I'm gonna write it down right here to solve for T. And I'm gonna plug in the variables appropriately. I'm gonna do that down here. So instead of me riding accident rate T because time is when I'm looking for equals Negative B So negative 5.36 plus or minus. And I'll come back to that in a minute. The square root of B squared 5.36 squared minus four times 4.9. That's the a turn. Times see, which is six. The entire top value gets divided by two times a, which ended up, which was really, uh, 9.81 right. But I just rated at its 2.2 times 4.9. Now, I think it's important to go through steps, maybe figure out the under the radical and then figure out the whole top value and then divide by the bottom value. Um, I think it will be up to you if you want to do that. So that's what I'm gonna do, though. I'm gonna rate negative 5.36 plus or minus. And I want to figure out my radical, and I'm gonna take the whole thing. Divide by. Well, two times 4.9 is 9.8. So it's were 89.8. So let's get this top number. So I'm gonna do five point 36 squared minus four times 4.9 times six. I think I've got something wrong here catching myself because I can't have an edit. Negative radical, can I? That's a problem. So what did I do wrong? Oh, who who? I know what I did wrong. I'm because some of you have been yelling at me this whole time. It's where you wish you could jump in and get into my head. I'm not gonna do the whole thing. Gonna look at all we appear. What's a negative? 9.81 Look here. Do I have the negative here? No, that means my name. My A term over here away. Okay. A team over here is negative, which means it's negative inside here. Which means it's negative down here. Oh, man. Oh, man. Oh, man. And this is why the quad formulas tough. This is why students avoid it. Um, you should avoid it. You should just try to make sure you catch these errors like I just did. Let's recap. Why did I know there was an error? Can I have a radical? Negative? No. You're never going to get a negative under the radical. It's not gonna happen. Don't just magically change a positive. Okay? Again, It's negative because it's negative over here. And I didn't carry that negative in when I worked down. Even though I schooled you and caution you to make your negative proper over here, I totally gaffed up over here. So rewind. Let's do this again. 5.36 squared minus four times minus 4.9 times six. Yeah, that looks a little bit better. 146.33 men get. I'm gonna take the square root now. Second square. Second answer 12 point around. That 12 point one's a plus or minus 12.1. So now what you gotta do is you take your negative 5.36 and add 12.1 and then divide by negative 9.8. That'll be your first answer. Then you do it all over. Good negative. 5.36 minus 2.1. Divide by negative 9.8. That'll be your second answers. You get to T values, so T one is equal to a well, let's do it. Negative 5.36 plus 12.1 Divide by a negative 9.8 and that's give me a negative answer. So let's give you the one I'm gonna end up throwing out mentee to let's do the whole same thing. Negative. 5.369 a. Subtract 12.1. You divide that by negative 9.8. That's giving me a positive value. That's what I want. Positive. 1.78 seconds. Cool. Now that I've got my time, that's the time it took to hit the target. So now that goes up here Minoans list. And now I can find my displacement. The displacement ends up being the easiest. This last step ends up being the easiest step. Because for this, all I've got to do is use my basic equation of average velocities displacement Over time, I'm gonna do this work right over here in this quadrant to just wrap this up kind of quickly. And I'm going to say average velocity is displacement over time. Displacement is what I'm looking at. So it's my average x times my tea. So it's the 4.5 times with 1.78 that I just found. Let's go ahead. Multiply this by 4.5. I'm gonna get a range of about eight meters. Ah, awesome. I know, I know, I know. It's a lot of work to get to this final problem of just eight meters. It is what it is. Um, sometimes I do a lot of work to find a good answer. So pause this. Rewatch it. Do whatever you have to do to get this down. You will have to do this likely, at least in a physics class. So hopefully hopefully this helps. Alright, that wraps this up for project emotion. Thank you