 Any questions on those homeworks? All right, let's see, so far we did a real quick one day review of rectilinear motion. Remember, this is strictly one dimensional motion. So even if we have a track on an object has to follow, which you think would be kind of like one dimension, you can either go this way on the track or that way on the track. That's not rectilinear motion because the track itself exists in two dimensions. So that will be what we'll look at now which is curvilinear motion. So what we looked at with the rectilinear motion, that was strictly one D. It could only go left or it could go right as if it was on a train track and then the train track that was a concern was very, very straight. No turns, no elevation changes, just out and back, nothing else. Now we'll do a little bit more with it. We'll do it in a couple different ways. We'll start with the usual that we're most familiar with. We'll look at it from Cartesian coordinates. Not a terribly big deal, that's what we did in physics one. That's kind of how people think about the world anyway for the most part. Everybody orients of themselves at least maybe in their mind's eye with the north and south and east and west and then no matter what you need to do you do it within that two dimensional plane defined by those ordinal directions. We'll also look at a different coordinate system called normal tangential. This'll be particularly useful for a certain type of sort of regular and controlled 2D motion as we'll see. And then we'll briefly touch on polar coordinates which is another way to look at 2D motion that for certain problems can be quite an advantage. All right, so we'll begin with the Cartesian business. As usual, as it did with the rectilinear motion it all depends upon some defined, agreed upon origin. Some place from where we measure everything. Where that is, that doesn't really matter. Whatever's of particular use to a problem that'll be good enough at that origin then we'll stick a simple Cartesian coordinate system. Typically, though we do not need to, we'll put x horizontal to the left and right, y vertically up and down and unless it lends itself to a particular problem to do otherwise we'll even take right to be positive, up to be positive. But we could do anything we want as we remember the origin as well as the orientation of the coordinate system is arbitrary and whatever's of most use to us in the problem. So once we've established a coordinate system from where we can measure things, we can, well then, measure things. So here's an object at point one. We might then give it a position vector that we might call R1 that locates it at some time. Maybe we even want to be so creative to just call that at time T1. Now we're looking again at curvilinear motion. So we're gonna exist in this 2D plane of the border of in your paper or it looks and so there may or may not be some path that's defined. The object has to go from one place to another in some way. So it's certainly gonna follow a path. We may or may not know what that path is, but we can take it as that and it'd be as simple as that. Once we've done that, we can then define from that the components of that vector. So this might, let's see, might call this R1 in the x direction. This we might call R1 in the y direction. And then we'll be able to use all of our regular trigonometry to define all of this in any way that we want. And so this R1 vector, we might call as made up of two orthogonal vector components. So we could do that with it. I'm not a real fond, not a real fan of using too many levels of subscripts if we can avoid it. I would hate to go much farther than two, but if I can even avoid two layers of subscripts or even one layer of subscripts, I'm happy to. So there's other things we could call this that would be just as instructive. We might call it, well, if this is R in the x direction in the x direction, we might just call it x. Maybe if we're worried about a couple different places, we'll call it one. That'll save us a little bit, but again, it's all just notation. That the sole purpose of doing that would be to save us one level of subscript. We might want to call that a vector component itself, a vector itself, or we might also use the unit vector notation with which I hope you're all familiar, looks familiar. Everybody remember it for the most part? Alex, you remember, we did that in physics one, right? The unit notation. And then we might call this and y1, because we're talking about the first position again. Just make sure that your i's and your j's look different. It's very common for some students to not make them look visibly different enough and then you and I run into trouble going over this work. So we've got all kinds of possibilities like that. All right, other stuff that might be going on. Let me just clean this up a little bit because I'm gonna need some room with it. So here's our arbitrarily chosen coordinate system from which we define everything. If we want to, we can remind ourselves that we've got unit vectors that define themselves with that coordinate system. Remember, those are used just to give scalar components a directional quantity to it. All right, so we've got some path we're following here in an object at some point, r1, we might follow it. And at that instant, the object following that path will have some velocity which we might call v1. It must be, at any instant, by definition of the path, it must be tangential to the path. If there's any component to the velocity, that's not tangential to the path, that's orthogonally perpendicular to the path, then that component of the velocity is gonna take it off the path. It must. It's no different than you driving down a highway. If you have any component of the velocity, of your velocity sideways to the direction the road's running, you're gonna leave the roadway. Some of you like to do that on purpose. Others don't. So we can relate these. This velocity is the time rate of change of the position vector we'd like. We can use the dot notation. But it's also the time rate of change of p to the component direction. So maybe we'll call it x1.i plus y1.j. Staying away from the r1x notation, which can in itself be a little bit of a pain. But all of these mean the same thing if we want to. We mean the vx plus vyj and all of the other possibilities that can go with it. Any of those are fine, recognizable. You might want to sort of zero in one in particular and stick with it. Some are more useful than others in certain problems, as we'll see shortly here, I think. But you're welcome to use any one of the different types of different possibilities and notations or the very same thing in whatever makes you comfortable with it. All right, some time later, we might be at some other spot, point two, and there'd be a position vector. I'm not going to draw that because the things would get a little too crowded if I do. And we have a tangential velocity at that instant. Maybe we'll call it v2. Maybe the magnitude of v1 and v2 are the same. Clearly the directions are different. Maybe the magnitudes are the same. Maybe they're not. It depends entirely on whatever problem we're doing. But once we have that, of course, we can define an average acceleration just like we did in physics one. Define as the change in the velocity vector with time. Delta t being whatever time it took us to go from one to two. But that velocity vector is very strictly v2 minus v1 over that delta t where this is a full vector equation. It must be treated as such, it must be kept as such. It makes no sense if we have equal signs in here, if one parts a vector and the other part isn't. So if we have a vector equation in one part, we have a vector equation in all parts. Well, we can look and see just exactly what that means. Here's v2, we'll add to it the negative of v1. Let's see, v1's like that, the negative of v1 looks like that. Then we have a change in velocity vector. That must be parallel to and in the same direction as the average acceleration vector. So now we know that in between points one and point two, we have an average acceleration vector in the same direction as the change in velocity vector, just like we would have done before. I do negative v1 because I personally, when I'm subtracting vectors get confused of what tail goes to what tail and then which direction the resultant goes. And if I do v2 minus v1 drawn this way, I just never mess up and I always get the right direction for delta v. That's just maybe my mental shortcoming. Maybe you can subtract vectors a little less haphazardly than I can, you're welcome to detail. That's just one of my little mental foibles. You can Google that word later. Of course, we're gonna be interested in the instantaneous acceleration, which is the acceleration at any one instant as this time period goes down to a single second time. And that then of course is just the straight time derivative of the velocity vector with all the other notations. None of this is any different, or at least the what's up here really, none of this is different than what we had in the rectilinear motion. The fact that we've gone from one dimension to two has not really changed things greatly, other than the fact that before our direction in one dimension is either a plus or a minus. Now in this case we have the full 2D flavor the possibility of a vector in 2D space. Other possibilities for this, we can break it into its component directions just like we could with the other one and using any of the notation that you so choose and all the variations there of BX dot BY dot X double dot IY double dot J, whatever of those notations looks the yummiest to you. CJ got a favorite? All right, the acceleration vector itself, of course, can also have the X and the Y components and we'll have to take care of all those as we go. Even though that the velocity vector will never have any normal component to the path, it can only be tangential to the path that's not true of the acceleration vector as we've simply seen here. There is a component that's along the path as well as a component that's across the path. So there's more stuff that comes into the acceleration than what we had before in 1D motion. All right, so we're gonna look at a particular type of curvilinear motion just like we did in physics one. We're gonna revisit this and that's the idea of projectile motion. Projectile motion is, we're gonna look at it, is a free fall problem. That should mean one thing, most certainly to all my students who are from physics one and for those of you I didn't have from physics one, we'll remind you of it. Just may not have been put so succinctly so explicitly as I tried to do for my physics one students. Jake, you nodded your head, that word free fall means one thing when you hear it. Acceleration due to gravity is constant? That's sort of a subset of the one thing you need to know. You need to know something actually a little bit more in general and then that falls from it. No, I mean it is, yes, the acceleration is constant but you need something more than that. What? You were saying, wait, yeah. Gravity is the only force in the problem. We neglect air resistance as you've seen in lots of these problems. They all say even something in the problem like neglecting air resistance and then they go ahead and lay off the rest of the problem. We will look briefly in this a little bit later in the term at what we do to include air resistance in a fairly simple way. That's an extra exercise in it. All right, gravity being the only force then gravity, the gravitational acceleration is the only acceleration we have. It's in the wide erection and unless you'd rather, we'll take the downward direction as the negative direction. You may have any severe protests? Bob, even you're okay with that one? Okay, so we'll arbitrarily then put in a minus sign. That's just the agreement between us that we'll take downward as the negative direction. We don't have to. It's an arbitrary decision but like good minions, you do what I say. And I said, show your belt followed along with it. And it'll have a field strength of G. We'll take G to be either in general 9.81 meters per second squared or 32.2 feet per second squared and less told otherwise than one of the problems. We did look at chapter one a little bit of what happened with the different altitudes and different places on the globe and other kinds of things that can affect that. But unless it's specifically said otherwise, we'll take those to be the values in here. Notice, I hope G itself does not have a negative sign on it. This is the gravitational field strength. Most of you are in physics three now. So you're gonna start, if you haven't already, you're gonna start talking about field strengths you'll talk about the electric field, the magnetic field and the like. So you'll know a little bit about that. It just tells us if we had a small mass and let go of it in the gravitational field, it would achieve this acceleration. If we go to some other place in the planet like we looked at some of our altitude and let go of that same mass, it would have a slightly different acceleration as we looked at in chapter one. This negative sign, we arbitrarily impose. So it goes in front of the G once we've made our coordinate decision, but G itself never has a negative sign in it because that's an arbitrary imposed by us whereas these values are not. All right, so if we wanna be a little bit more explicit we could even make this a vector, make sure that we've got it right. We know then what the acceleration is. We'll take this to be as well, a constant. So now anytime we're looking at some kind of projectile motion, it doesn't matter where we start, it doesn't matter what our initial velocity is, it doesn't matter where we land. All of it shakes out to be exactly the same for any projectile motion problem. It all depends on some initial velocity which may or may not have some angle to it, doesn't matter, I'll, I have to draw something, so I'll draw that, but it doesn't matter what theta is, theta can be anything from the zero to 360 and all of this shakes out to be exactly the same type of thing. In the presence of a gravitational field, an object with some kind of velocity, something like that will tend to follow a parabolic path through 2D space, realistic looking in it, makes it feel like you're there, actually kind of looks like you're picking off a cliff, that's why I don't use the yellow chalk. But any projectile problem we look at, whether or not the launch site is above the landing site, whether or not it's to the left or to the right of the landing site, whether or not there's a zero theta launch angle or the 132 degree, none of it matters, it all turns out to be exactly the same type of problem, it's just a matter of us getting a projectile motion problem, looking at what the given quantities are and then putting them all into the exact same equations. Those being, split between the horizontal motion and the vertical motion. There's two different things going on in the two directions. Actually, actually the same things going on in the horizontal as in the vertical motion, it's just that because of what the values are, things get greatly simplified on a horizontal side. On the horizontal side, the acceleration is constant. As such, the constant acceleration equations apply. By the way, I did put up the constant acceleration sheet on Angel, I don't know if you asked for it, right? I think there's another copy of that. So if you go to Angel, the constant acceleration equations are there. We do have constant acceleration in the horizontal side, it happens to be zero. So any of those constant acceleration equations apply, but any of them that have A in them, you have to put in a value of A equals zero. What that means is we're only left with a single, really one single horizontal equation, one single constant acceleration equation. And that's because, let's see, if the acceleration in the x direction is zero, then the velocity in the x direction is constant. Of course, this isn't true in the vertical direction because we have this constant downward acceleration. So the velocity in the y direction is not constant. And it's the, those two conditions, constant horizontal velocity, non-constant vertical velocity, but with constant acceleration, it causes the parabolic motion of the standard projectile through space. Since the velocity is constant, then that's the only constant acceleration equation there is at our disposal. None of the others are useful. If you go to any of the other constant acceleration equations, put in A equals zero, then you get something ordinary and identical, like V one equals V two, which of course it does, there's the velocity's constant and that's no help as one of the equations. So that's all we've got on the vertical side. If you'd rather, and this is one time when I prefer to do this, you can use the dot notation. I like to do this because then it's a reminder that on the horizontal side, we shouldn't see anything but an x. There should be no y's over here. If you remember, we separate the vertical and the horizontal with what? A semipermeable membrane. Yeah, you didn't hear this before about the semipermeable membrane? Sounds a lot better. Yeah, it is, it's very important. It's absolutely crucial. As we'll get to, we'll tell you why it's semipermeable membrane in a second. Also because it's constant, if we have it as given, we can then also establish that in terms of the initial velocity. Of course those are both constant. Once the object's been launched, you can't change the initial velocity and the angle of launch because it's already in the air. So of course those two things are constant. But that's the entire setup for the horizontal side of the equation. On the vertical side, we've got this business that we have a non-zero acceleration. If you wish, you can use the dot notation. But then the velocity is not constant and it will be a function of the initial horizontal velocity plus any change in that velocity due to the acceleration. That actually is just one of the constant acceleration equations. This is that the acceleration is defined as the change in the velocity with time. That's all that equation is. There's nothing magic to it. It's just one of the constant accelerators in places. Jake, did you have a hand up? Yeah, this is probably just derived from the constant acceleration problems. But isn't the horizontal, isn't the delta x equal to like the x t? Is that, or are those like another? Delta x is equal to vx t, right there. It's the only constant acceleration equation that's left. Remember, one of the constant acceleration equations was the average velocity is the change in position divided by the time. All the other constant acceleration equations drop out. As do all other constant a acceleration, constant a equations apply. So there's no need to write them down. You can write them down. You may want to put them in the dot formation because once again, this visually emphasizes that on the vertical side, there are only allowed things with the letter y in them. On the horizontal side, we only allow things with the letter x in them. We strictly practice in this class segregation. It's an absolute necessity in this curvilinear motion that we keep the horizontal motion separate from the vertical motion. They're both happening at the same time, but they do not interfere with each other in any way. We just simply add them together. As a semi-permeable membrane, that implies that some things can cross from one side to the other. And there are three things that can be found on either side of the equation, either side of the semi-permeable membrane. Only three things can pass through that membrane. That's why it's semi-permeable. These two things are happening at the same time. So any time value you use over here, you use over here for the same part of the motion. So we'll go back and forth the time. Semi-permeable membrane. It lets some stuff go through, but not others. No x's can go that way, no y's can come this way. Jake, one of the others? The angle? The angle, because that can play a part in either one of those because this establishes the entire velocity in the horizontal direction. It also establishes the initial velocity for the vertical motion. And the third thing that can pass back and forth between this, through this semi-permeable membrane. What? Units? Oh no, there's always units. The units don't go anywhere. The units, if you bring over a value for that, it's gotta have the units with it automatically. They're inseparable. DJ? The initial velocity of it. Magnitude of the launch velocity itself. Any of the other constant acceleration equations apply. So no need to little any of them. Once you realize what you've been given here, then you solve that side like any other constant acceleration equation. Nothing else matters. This is true for any launch velocity. No matter what the angle, no matter what the magnitude, even theta could be 90 degrees, which is one of those problems where the objects go straight up and straight down and it goes back to a rectilinear motion problem. It doesn't matter if theta is greater than 90 or even greater than 180. Well, if it's greater than 90, we just have a picture that's been flipped around and there's nothing to worry about it. It's just, it's a very same problem that the class on the other side of the wall is seeing in our alternate universe over there. Where your alpha e goes out. It also doesn't matter if we happen to have a launch velocity that's got it, that's downward and we follow that kind of projectile. It doesn't matter. All of the problems are solved in the same way. So let's do one. Doesn't mean the problems aren't phrased differently and don't ask for different things and have different things given, but they're all done in exactly the same way. All right, let's start out with a real straightforward one. Here's our little cliff. Huh? That's okay, I'll play the dig back. In fact, I think I'll just get a directional mic and just focus it right there. We'd say all you want, I'll catch it. All right. 110 meter high cliff. In fact, I did this just over the weekend in my backyard because we have that cliff in our yard. These are, we'll take in this case, state of the 90 degrees just to get it started. 10 meters per second. So stand at the top of the cliff. You throw your little brother straight up in the air. Remember, we are not concerned in this case since it's never appeared anywhere. It can't be a concern with the mass of whatever it is that is the object. We're not talking here strictly about what the forces are. Just the force of gravity is the only one. All right, simple problem. Find the maximum height it travels above the base of the cliff. These saws still apply. It's just in this case, the horizontal velocity is zero. So this entire side drops out. There's nothing to do with the horizontal side. It's all gone since both the acceleration and the velocity are zero. All we've got to work with is that side, the vertical side. It's a constant acceleration problem, which if you remember means three things are given. One thing is not, and that's how you decide which of the constant acceleration equations to use. What three things are given in terms of the vertical motion? Since we've already come to understand the horizontal motion is completely out of the picture. You've got the initial vertical velocity given as 15 meters per second. What else is given? Sorry, Pat? Position. As is, what is the initial position? Right, so do it the way you said. Initial position. You said about three things, I think. You started arguing with yourself. You could say 110 or you could say zero. You just have to put it down. Which is right. What, DJ? What's the right thing to put down for the initial position? Depends on where you put your origin. Where's your origin? What did I tell you about the origin? Where's it got to be? Wherever you want it. So, Pat, you spoke up first. It's arbitrary. Just put it somewhere. Agree to it and then don't move it. All right, you're just gonna say y equals zero is there. So our initial position then is zero. And in fact, anywhere it is from there will be just the part that would need since Pat conveniently chose the initial position as the our origin. So that just makes things a little bit simpler. What's the third thing that's given? Pat didn't say so, but I'm assuming there that down from there is negative, up from there is positive. Doesn't have to be, but we're all pretty used to that. So, three things given. We need to find, find what? Find, we need to find the maximum height. Find delta y. Little bit of a trouble here if we're not careful. This initial position is not actually part of the constant acceleration equations. Delta y is part of the equation. Pat's conveniently chosen them for us, but we have not been given the third part we need for the maximum acceleration equations. So we need one more piece of information because nowhere in the constant acceleration equations is y zero up here. We need the information that defines this point of maximum height. So that's actually our third given that the velocity will call it two, I guess it'll fall from there if we let it. Y two is zero. We want to find the height where the velocity itself is zero. So those are the three things that we want to use for the constant acceleration equation. Only one equation has those three things, so that's one to use. Check your tattoo. Good work, you got it already? What's so funny? All right, let's do it like that. Thank you, that's how to do that. Frank, you got your tattoo? Got the constant, do you guys have your constant acceleration sheets? Maybe not your tattoo. It's at home, I gotta drive it. You can write it on the board right off the side where they can't see that camera. Right. Well, it's like reference to the rest of the semester. I don't know if I have a copy of this or any of it. That's pretty good. Those are just like the, those are formative sheets or those two sets. Is that? Go to the dynamics part under resources, under useful links. There will be the constant acceleration equation. There we go. There's your constant acceleration equations. And if you go to angel, this exact page will show up there. So we have a problem with initial velocity, final velocity, acceleration, and we're trying to find a position. So we have initial velocity, final velocity, acceleration, and we're trying to find position. So that's the only equation that'll work directly right there. Three things you know, three things we didn't know. So there's only one equation that will work for us. So doesn't that seem useful as a tattoo, Frank? Just to have it right there in your forearm or something? I have it on my, the inside of my eyelids. Is that blank? I read it, there it is. I've got every course I've ever taken to there. So we know v2 squared equals v0 squared. We're using the same notation we had over here. Minus 2g delta y. And then we can solve that for delta y. Because that's what we're looking for. Did it work? Is that what you did? Colin? Okay. What'd you get? 11.5. You shouldn't have, as long as you're careful with everything else, you shouldn't have any danger about the possibility of a minus or a plus sign is there. Sometimes we do with the time if we have to use these equations, if we have a quadratic and time, you have a choice you need to make of the plus or the minus sign. In fact, any questions on that for a clear? Let's do the next part of this problem. Figure out the time when it hits, the base of the, let me get that one fixed too. It's got just enough, a little tiny bit of horizontal motion, just enough to make it clear the lip of the clip when it comes down. Find the time when it then hits the bottom of the clip ignoring the fact that if it really is thrown up vertically, it will hit the clip on the way back down. Well, you're leaning over the clip a little bit when you throw it. So, here's your break. Again, three things you know, one thing you don't. Three things you know. One of the three things. Initial, that's still the same problem there. We still have the same initial velocity. Second thing, not in any order, but you've got three things you know. Acceleration is minus g, we've already established that down is negative. So, we have to continue that. What else is known for this part of it? Do we know that second velocity? Why don't we know the second velocity? We just found it out. That's a completely different point. That's up here, so maybe we'll call this point three, just for labeling purposes. So, what's the third thing we know? Remember, it's got to be one of these possibilities. Change in height is, what's the change in height? Negative 110, but didn't it go up 11.5 meters? How do we handle that? As long as we're interested in the change in height from here, the original point to here, it doesn't matter what it did in between. In this case then, we have a delta, we could put a delta y from zero to two. This is from zero to three. And it's crucial that we make note of the fact that it finished 110 meters below where it started. Doesn't matter what else it did in between. When we have separate problems, separate places, judicious choice of notation can make problems a little bit simpler. Which equation involves these three things and the time? Initial velocity, we have the distance of travel, we start to finish with acceleration and we have the time there. So, the equation we want, the quadratic solver on there, quadratic equations are, no, in any budget. I like, I only got the quadratic equation solver on there. You can remember the quadratic equation too. That'd make a darn good tattoo. What do we? Why would you do y as a measure? Shouldn't, let's see. Let's go ahead and fill out all the pieces on our quadratic. Let's see, we're working right in meters. That's the minus one half AT squared plus VIT. And if we bring that over, it's plus 110. So that's the quadratic equation we need to solve. That's C, right? So, did you make a mistake, do we? Yeah, I forgot that's what. Forgot what? Negative on the G. Oh, yeah, be careful. Didn't work. Well, hang on, let's do one problem at a time. Right now, we've got laid out doing it from zero to three. You could do it from two to three, yeah. It's just these, the values might change. Now you have a different initial velocity for that time period. Remember that we're talking about initial and final velocities. We're talking about a particular chosen time period. Pardon me? But you should get the same answer. So we have quadratic in time, so we should get, as with most quadratic equations, two answers. The two answers, you gotta look at them both, because until you have both answers written down, both roots written down, how are you gonna know which one's right and you didn't just have to pick one to start with that wasn't right? You need to look at both answers. It's not that difficult programming these things. Even I've done it. Two roots. Remember, there could be two, there could be one, there could be none. But for this type of problem, I can't imagine what would have happened if we first had no roots. But we could certainly have two roots. Got two, Alex? Do you have two? You check. Smile at her. Just one root. Colin, what do you have? Two seconds. What do we have? 6.5 seconds, I think I heard. And 3.4 seconds. Now, surely obviously only one makes any sense in this case. That being the 6.5 seconds. Is there any meaning to the minus 3.4 seconds? The minus 3.4 seconds tells us, well there's two different ways to look at it. It's the time it would have taken for a projectile to start down here, go up to this point, pass that point as long as it passed with the same velocity, 15 meters per second passing that point, it would still go to the same height and then still take an extra 6.5 seconds to come down. So it would have taken 3.4 seconds to get to here if it went with that same velocity and then it would have taken 6.5 seconds to come through. What's the initial velocity down here that would give us that? I don't know, we don't have that number. Or you can look at it as the time it would take if we shot downward with 15 meters per second how long it would take to get to here. So there is physical significance to that, it's just for this problem, we're not concerned. All right, we're going to change things a little bit. We'll keep that same launch velocity, sorry, we'll keep that same launch speed, we're just gonna knock it over a couple degrees. So still 15 meters per second, but now it's 78 degrees. So I want you to find a couple things, find where it lands and the velocity at impact. Find where it lands, commonly known as the range of the velocity at impact. The Von Angel, a projectile motion summary, just the very same things we've been going over for any initial velocity, the things that apply. Again, it's the same business on the vertical side. It's constant acceleration, three things apply, three things are given, one thing isn't on the horizontal side, it's a constant velocity problem, which is a type of constant acceleration problem, just a very, very simple one side. Three things you know, one thing you don't, it's on the side, only one possibility, that's the given velocity. To figure out the range, we're gonna need that velocity, but we also need the amount of time it was in the air. How are we gonna figure out the amount of time it was in the air before it landed in this problem? We need the time to go with the velocity to find the range, how far down range it was when it landed. How are we gonna find this time? We're gonna find this time so we can find out that range. The velocity in the x direction is no trouble, that's all I'm given just to figure it out. But how long, what time it was in the x direction, traveling in the x direction, we need to find this range, how can we find that? Without using the vertical? Who said without using the vertical? In fact, remember I said that time is one of those things that can cross the membrane. So that's where it's going to need to come from. But it's even easier than that. How do we get the time that it took before it landed? Do it exactly the same way we did it before. Three things you're given, you're given the initial velocity in the y direction, you're given the change in position in the y direction and the acceleration in the y direction, you can find. So your question, should it be the same time that we had in the problem we just did, the 6.5 seconds? Is this the same initial velocity we had with the problem we just did? The launch speed is the same, that y component, the vertical component is different now, or a vertical component there, different. V o is the same, but y dot o is not the same. That's five minutes here to fix this one up. You got it? Who's the checklist? You know what, still not talking to anybody. Furthering and fostering the notion that engineers are socially inept. Frank, who'd you talk to? Do you ever met anybody in the interview site? Is it going to be in the air longer or shorter than before? Short, there's less vertical velocity now and that's what determines how long it's in the air. So that's slightly less long than it was in the air. Anybody else get the time? 6.47 seconds in the air? That's what they got. Once you've got the time in the air which you get from the vertical side, then you can bring that over. Anybody have that yet? 0.16 meters. We'll call it 20.2. I think you've got the impact velocity, the speed with which it hit, actually the velocity with which it hit at the base of the cliff. They figure that part out. Did you get the range okay? How do we get that impact velocity? It's made up of two parts, the F in the X direction, the F in the Y direction. How do you get the F in the X direction? Yeah, remember there's no change in speed in the X direction. So we've already, we haven't actually calculated it. I don't have it on the board but you've already done that with the range. How about that final velocity in the Y direction? How do you find that? Constant acceleration equations. You know the initial Y velocity. You know the Y acceleration. You know the Y changing position. So it's a constant acceleration problem just like any other. Any questions for Dan now? Bobby okay? Yeah, for the last part. To find the Y component, you look at purely the vertical side. You've got the initial velocity. You've got the acceleration, it's gravity, downward. And you've got the change in position in the Y direction. So those three things will give you the final velocity in the Y direction. You add that to the constant velocity in the X direction and you're all set.