So in this section here, section 2.8, we're going to consider uh free a 3D folding object and first we start with kinematic equations. So recall that recall in the previous session We have the four kinematic equations. Hey, so let let me remind you that what are the four. So the first one was um we start with the the v, right? So vf final equals v initial plus ax * t first equation. Second equation xfal x initial plus 12 v initial plus vf final times the time. Right? What's what was the third one? Third one is also for X. X final equals X initial then plus V initial * time then plus 12* the acceleration time the time square. And the last equation we have is Val square = V initial square then plus 2 * the acceleration in the X direction then times the displacement in the X direction. Right? And last time we say that we have an object say a ball and I toast this ball vertically upward from a certain location. So that means our dimension in interest actually is for this section. So now let's the previous four equations was for section seven right and those four equations will be provided on the on the exam paper. Right? I only will provide those four equations. All other equations you need for the for the exams, you should write write those equations down on the no car, right? Each exam will permit me one no car three by five inches. You can handr write things on the no car. But those things must be handwritten by you alone, not by some other people. And we don't accept uh printed materials, right, on the no car. But now let's move on. So now we want to change those four equations into equations for the y dimension. So assume that this is your positive yaxis and this is zero and this pose was toasted from a certain initial location y initial somewhere and the velocity is called v initial right initial velocity. So now what is the direction of acceleration? This we talked about this last time already. Direction of acceleration is it up or down or to the right or to the left? Down. Hey, by real life experience, you know that acceleration is everywhere downwards. So it's we call it a y. Now it's no longer ax, we call it a y equals negative of 9.8 m/s squared. And this value we give it a name. We we call it G. So basically a Y we call it negative G right. So we use such a langu language here. Let me say that again. We assume a Y equals negative of a constant. And what is that constant? That constant G equals positive 9.8 m/s squared. So that means G is a positive constant. But we assume a Y to be negative G. We use such a language, right? In high school, you may or may not use this convention. But here for this course, all the instructors, we we use this convention, right? And the acceleration along the y direction is negative because positive y-axis is up. So acceleration points down. So that's why it's a negative acceleration, right? That negative sign simply means that acceleration points to the negative y direction, right? So then initial velocity that let's assume here could be positive but we could also have other initial velocities as well. So now what we need is to convert these equations into the equations in y dimension. So simply you do want to do the replacement right. So replace the v in x dimension into v in y dimension. Replace a x by a y replace x by y. Right. So after replacement now we are going to have the following equations. Right. So I'm going to write down the equations after replacement. So let me write those write down those equations here. So equations for section 2.8. If you read a textbook there's no no new derivation. We simply use those equations in the y dimension. So in this section here we have V. Okay. Here v means the y component or the v in the y dimension. So vf final equals v initial v initial. Then replace a y byative g again. Okay. We assume that a y equals g. That's our convention. So therefore this will become negative gt. Okay. So that's the first equation. Second equation we have is y final = y initial plus 12 v initial + vfal times time. And that's the second equation. Third equation. yf final equals y initial then plus v initial time then minus 12 gt². Right? So that's the third equation. So that the last equation reads uh vfal squared equals v initial squar then minus 2 * g and * uh y final minus y initial right so those are the four kinematic equations like like I said these equations are not new those are simply I have rewritten the previous four equations here into equations in the y dimension and I I had such a replacement here. A y= negative g and I got those four equations. When you work on those four equations, make sure that you realize that those equations are in the y dimension, right? And this v initial was the initial velocity. And if you see vf final, that means the the final velocity. And like I said, those initial and final doesn't necessarily mean that ultimate initial and ultimate final. This V initial doesn't necessarily mean the beginning very beginning right could be any two points okay could be any two points during the course of motion and that V initial means the velocity at the earlier time and that V final means the velocity at the later time right so so let's just move on to uh some some uh practice questions This Um okay. So now let's uh move on from here. Apologize for the waiting. Apologize. Okay. So these sections here we have finished the theory. It's quite straightforward and let's look at this example here. This one is in the textbook. I believe it's this straightforward although the some of the phones could be quite small. So the this question you don't need to write everything down. I mean but of course important equations you should write those down. But this equation is a very t sorry this question is a very typical example for in this for this section here. So non if you have somebody who stands on the top of the roof of a building and this person he toes uh a stone vertically upward initial velocity is given 20 m/s and straight upward that. So that means once the stone leaves the hand of the person then there is only one force exerted on the the stone right suppose I don't have a stone I have a bowl here once so we our init very initial time uh we consider at the instant where this stone just leaves the hand of the person that's the our initial time so don't need to you don't need to worry about what's the force exerted by the hand of the person on the stone that's irrelevant. I mean irrelevant to the solution to this problem, right? You don't need to worry about what's the what's the force of the thrust that this person exerted on the stone. It does it doesn't matter but you just being given the very initial time when the stone just leaves the hand of the person and that initial velocity is given and and that's all. Yeah. And then what's the force exerted on the stone will be force of gravity. Right? So after this moment it would just be force of gravity. So that means is acceleration constant afterwards. Yes. So that would take that this constant value of acceleration. So that means you have those equations in mind. Now we can solve the problem here. So now the stone is launched at a certain initial location. So you know the initial location as well right? So that that initial y can assume and now let's move on here. So they have set up a coordinate system for you right. So now they they can they consider such a a scenario here. They consider the very initial location. They call it zero. They prefer to call it zero. They prefer to call the at location A here when the stone is just being launched they call the Y at location A equals zero. So that's their convention. So that means when the stone drop all the way to the bottom of like a to the ground then that Y will become negative. Right? So but that's just a choice a choice of mathematics uh or algebra from from the authors. Right? You can have a different choice but now let's just follow their choice to see what kind of result they can achieve. Right? So part a of a question at t equals z right? So determine so now when t equals z the stone leaves the the some this person's hand at a certain position a and determine the time at which the stone reaches the maximum height. Right? So that's what we want here. So now let's so can you read the question here? It's pretty large here. Right? So now let's see now let's do it first do it symbolically and then we'll move on to some some other consideration from here. Let's erase this. So first let's label all the coordinate system. So this we call it y equals z. Okay. So y a equals z. So this is our location a and we have a initial speed very initial speed. We can call it b initial which is given. Now let's first address the problem here and we want to get the maximum height actually right. So here we know the initial speed which is 20 m/s and then the build height of a building we call it say capital H is 50 uh mters that's what we know of so now they want to get the maximum height so cons consider your real life experience when you toss this stone vertically vertically upward so when this stone reaches the maximum height then what What is the velocity there at the maximum height? Zero. Okay, we talked about this last time. That's great. So that maximum height we call that location B here and let let's consider a schematic view here. So suppose this is our location A, right? And when this stone is toes vertically upward, you will reach the maximum height and it will drop again. So that maximum height, we call it location B. Assume that, right? And that version B. Now let's see what we know and what we do not know. What we know is actually VB equals zero, right? That's what we know. That's a very good information for us. So now let's identify the knowns and unknowns. So here you know that V initial equals 20 m/s. And that V initial you can call that VA for example, right? That V initial is is just just VA because now they don't want to use A B C D to consider a problem. So let's call the V initial just call it VA and then VB equals zero and other things we may not know very explicitly but it's okay now let's choose the equation use right so and now what you consider is as follows if you want to choose the equation to use so as to help you solve the maximum height what what would you do the method is not unique right method is not unique now if you want to solve for maximum height. So now now you know the problem here, right? You know the problem very well right? Now and now don't don't look at the solution from the textbook and usually when we so so sometimes I don't really like to talk about textbook examples because solutions are all there and if you read the solution first you you lose the opportunity of uh critical thinking. So so let's not looking at the solution there and think about what we should do right. So if you want to determine the maximum height here then which equation would you use right? Which equation would you use? If you want to follow the test solution you you can try to use the first one right try to use the first one and see if you can solve a problem. Why why would why would they choose to use the first one? Because you know now you consider a as initial. So part a of a question part a part a of a question here I use lowerase a because I don't want to use that capital a again which can be confusing part a question here let's assume that we consider a point a as the initial point this is just my notation point a as the initial point and then point b assigned to be the final why did I do this because I want to be clear about which formula to use and try to use it correctly and as I've said this initial and final doesn't mean doesn't necessarily mean the ultimate initial or ultimate final but you just need to be sure that choose a period of time right and then that initial stage doesn't need to be the ultimate initial but we call it initial and in this case we call a point point a to be the initial point b is the final but actually we know that point b is not the ultimate final right it's just sum some some point we call you final point. So now let's use equation one right. So you can see that VB should be equal to VA and minus G * time and this time is actually TB actually right do you agree? Because this time if you we assume TA equals Z Y A= Z and you can say that TA equals Z. So the time that it reaches the the top we call that TB. TB is actually unknown but through this equation you can know TB right. So how would you know TB? So VB again don't write down the things on the on the screen right on the screen that's those are textbook notations you need to think critically so you don't want to just follow the textbook solution right so they their solution may not be the best may not because there could be multiple options right of solving a problem so now VB is 20 right so VB sorry VB is zero I'm sorry VB is zero meter per second right and VA is and minus G what is G 9.8 A right and times TB right and then you can solve for TB and now let's think about this once you solve for TB what what do what would you do next once you have solved for TB suppose now I have solved for TB and then what would you do next after you have gotten TB what's the goal maximum height so what is the maximum height Right. So, so you need to consider which equation to use. Right. Here you have ya equals zero. Maximum height could mean this location. We can call that location YB. Right? So, you solve for TB and then you try to obtain YB, right? And then you can consider which equation you will use, right? There are equations you can use as well, right? But but let let's consider the remaining thing. Okay. So you solve for TB then you need to use one equation to solve for the YB. So which equation would you use? So which equation would you choose next once you solve for TB? Yeah. So it's not unique, right? The the choice choices are not unique since here you know why you know VB equals zero. So you can choose the second one and you can also choose the third one. See which one is easier easier. Okay. Again the choices are not unique. So now let's let's look at next. First solve for TB. Right? So TB will be equal to the following uh 20 over 9.8. Right? So, so, so if you want to pay attention to the units, you will get seconds in the end because here you have meter/s and here you have meter/s squared on the on the denominator. So, eventually you will get a t in seconds. Okay. So now let's look at here if you follow the textbook procedure then mean so here we are actually following the test procedure here right but again you need to think independently. So then once you compute the numerical value you will get like 2.04 04 seconds. Yeah, if you uh keep three significant figures, you will get this value. But this is not our final answer, right? So our final answer will be getting the maximum height. How would you get the maximum height? You can either choose this or this, right? Let's choose the second one to see see if this works. So now choose the second equation. You have you have uh yb should be equal to y a then plus 12 v a + vb and times the time tb right now know everything y so but we want to get yb so y a equals zero right and here you have uh 12 va is 20 m/s right 20 m/s and then plus VB is zero right VB is zero uh meter per second but time is time right and time is this 2.04 04 seconds. That's your computation, right? So now you know that in in the end your SI unit will be in in terms of meters, right? So here I should separate them like this, right? So this question is actually not not very difficult. We can see that it looks like a pretty straightforward, right? So now uh okay. So part a of the question we we have already finished is here, right? So part A of the question they only want want you to obtain the time that at which it m reaches the maximum height. So that's part A of the question. So part B of a question here. Sorry this question is actually part B. So part B of the question you want to actually get the next maximum height right find the maximum height. So you find YB to be this. And then after you finish the computation you will see that the the value will be will be about this value here 20 uh 04 meters something like this so that will be the answer now you can notice that the test they chose a different method right they chose a different method so they chose equation three but I use equation you can check you you actually will get the same final answer right so any question up to this Okay, sorry I move on too fast. So because the let me say again let me remind you that part A of the question you you just need to determine the time at which the stone reaches the maximum height. So, so you obtain a time for part A and then part B of the question they want you to obtain the maximum height here right so that's we have finished part A and part B but now let me ask you is it possible that you get the answer for part B without knowing part A yes the final answer be that okay so so actually You you see the statement right above its initial position if you call this initial position zero then that would just be this answer right that's a good question so the on the test it will be clear that which which height you are looking for here they assume that initial location this at the roof it's y equals z that's the initial position here so now my question is as follows is it possible to get the answer of part b directly without knowing the answer of par a par a is the time right without knowing time is it possible to get the answer to power B directly if you think it's possible please raise your hand okay so if you think it's not possible please raise your hand okay so it's actually possible right how how would you get get the uh final position directly without knowing time which equation would you use number four okay can see that number four you don't have time explicity so let's see if we could use number four to achieve the answer. Right? So here I need some space. So now I think let's say that let me erase this. So when I solve this kind of problem, I slow down intentionally, right? Because actually doesn't take doesn't take too long to solve this problem. But usually we we would like to think when we solve such kind of problem, right? And this this this example is actually very good. I mean it help people think, right? But then you move on to work on the homework problems then then you'll be fine. I mean this section is not difficult and not not really a difficult section. So now let's consider part B again but with a alternative method. Okay. So they could they could say that part B and and and alternate method right if you do not know time let's let's use that equation here you have VB² equ= VA 2 minus 2 * G * YB minus Y A we know that YA equals zero right so therefore you can after substitution this is what you will it. So, VB we know it's zero. So, you have zero squared and then VA is 20 m/s. You square it minus 2 * and G. Make sure that here you have a negative sign, but G is a positive constant, right? It's 9.8 m/s squared. That's G. Then multiply by YB minus Z, right? Then you can solve for YB through this equation. You have YB should be equal to uh 20 m/s. You square it divide by 2 * 9.8 m/s squared, right? And then you play with the math, you get the same y the same yb actually you also get 20.4 m, right? So we can finish a calculation. So like I said the method of solving each question may not be unique that depending on problem may not be unique here. So now they want to part C sorry uh sorry I should I shouldn't put the answers here. Okay. So it's also always good to think but so determine the velocity of the stone when it returns to the height which it was thrown. So that means we assume that this location when it returns to the same height the location we call it location C. So now we wonder what is the okay so what is the uh velocity at this location here right so now to solve part C of a problem we need to make another choice so now now the initial is no longer sorry now the final is we make another choice for part C let me erase this right we want to make another choice for part C of the Huh? What is this? Okay. So now let's try to make this choice. Part C of the question. We are not interested in point B right now. Hey. So now we are interested in this initial. We can still set initial to be the very initial that that that's fine right and the final now we choose point C to be the final point right here this arrow means assignment I right so this I assign this uh you know what this means so now what what we're considering is as follows choose the equation that you can use the textbook let's read the textbook answer first although you don't need to write this down like at a very bit I mean this time you don't need to write this down this is one of the possible solutions right they know that they want to obtain y at look so vc want to obtain VC so now the first thing you should consider is as follows use real life experience VA is upward but VC should it be upward or downward at location C here downwards right you will point downwards so that's first thing you should consider VC is downwards and then you do your computation. If you don't make a judgment first, then you can make a mistake. You can look at just read their answer. You don't need to write this down right now. Right? So they use equation four and choose A to be initial, C to be final. And then they can compute VC², right? Because VA is known, right? But VC square is 400, right? The value is 400. Then they take a square root. But be careful when they take a square root. There are two possible roots, right? Two possible solutions. One is positive 20. The R is negative 20 m/s. Why did they choose the negative? Because they know that the direction is downward. Right? The math itself will not tell you whether the final answer is positive or negative. You need to make a judgment from real life experience. So that that's one way of solving this problem. Right? But like I say, the way of solving this problem is actually not unique. Actually not unique. Right? So now let's use a different way. Okay, par a question. Now let's try to use the information from from part A which we have already erased already. Right? I have erased information from par. But now let's try to use let's say try to use equation one. Right? Try to use equation one and try to solve for try to solve for that uh vc. If you try to consider VA as the initial and and this here uh point C as final, you can say that VC should be equal to VA minus G * TC, right? But now the problem is as follows. I don't really know TC, right? I don't really know TC. But what we can consider is as follows. What what is VC? VC is what we want to get, right? So VC will be equal to uh 20 m/s then minus 9.8 m/s squared times the TC. Now let's think carefully. So this time was TA equals zero. Right? TB have has already been obtained from part A of the question. What is TB? TB is about 2.04 seconds. Right? 2.04 seconds. That's we got it from part A of the question. Now TC is something we also we want to know but actually we don't know. But now let me ask you this question here. What's the relation between TB and TC actually? Okay. So think about this. You toss this ball vertically upward. Right? It will take some time for it to ascend all the way to the top. then you will descend all the way to your to a location which is par like a like at the same location of as your hand here. So during the course of motion from A to B then from B to C the total time from A to C is called TC and when the time for it to ascend all the way to the top is called TB. Let me ask you I mean real life experience what is the relation between TB and TC? Okay. So, so I think many of you probably have such a sense here. If there's no air resistance, okay, if there's air resistance in principle, TC will just be twice TB, right? So, we assume no air resistance. So then we have such a relation here, right? So now you can say that okay, my TC is just twice TB. So the twice TB. So now you know how to uh write down TC here, right? So here you should have TC but you know that it's twice TB here. So that will be just so then this negative G multiply by TC and TC is just so multiply by two then times TB which is this 2.04 seconds. Okay. So this factor of two comes from the fact that TC equals twice DB. Okay, like like this thing here. And once you have this written down, you can also uh write uh figure out the final answer. The final answer will be like will be like like about this value here. So BC will be equal to negative of uh 20 m/s, right? You see it's negative, right? Because you know this is about this this 9.8 you can think that is about 10, right? So you have negative about something like 10 times 2 * 2 is about four. So it's something like around 10 * 4 and that will means it's -40. So 20 minus 40 is about -20 right without calculator. But now you can input those into your calculator you will see the same value here. So this VC equals negative 20 here right? So sometimes if you use this equation sometimes it's could be a little better because now you can see directly that VC is negative right you you won't miss that negative sign if you choose this method here right so again again method is not unique if you want to use this it's also good right but just be careful this question here right when you try to take the square root if you let me say again if you use the testable method I try to get VC square equals 400 m squared over second square. Right? When you take the square when you try to get VC is actually equal to the square root of this quantity. But then you have actually more precisely you should have positive or negative square root of 400 and so that would be positive -20 right meter/s but you should you should drop the positive answer and keep the negative one. So therefore BC equals -20 m/s, right? I mean this part is just the uh I just restate the the this this solution here from the textbook but just be be careful. Okay, then they owe me some step. You should consider VC to be plus or minus right square root of 400 and that will be plus or minus 20 and then you choose the negative solution, right? So that's it. But you can choose right like I provided this solution and the textbook provided another. Again critical thinking is very important right because on an actual test they will not tell you which one to choose. So now find it. Yes. So it possible to solve this question. Oh yeah that's also possible. But but also be careful if you want to choose the velocity from from point B. The equation is not really the same. Okay. So like so let me readress the question here. So the one person asked which is a good question. Can I choose B to be initial and C as final? Yes. But be careful when you choose B as initial then here you should have here what the equation you have here is not this time is not really TC. It's actually when you choose right when you choose B as initial because here we assume initial time to be zero. Right? But if you would rather choose B to be initial C as a final then this time is actually the time you put in your equation that will become uh the following will become VC equals VB then minus minus G. Then you here you will have TC minus TB. So we need to be very careful if you want to choose B as the initial because B the time TB is not zero. Right here when you use this equation that's a great question right when you use this equation you assume initial time to be zero in principle this this part of equation you should write it as t minus0 something like this but we omitted this minus0 so that's why it became this equation but if your t initial is not zero then you should put your t initial here so this in principle this means t final minus t initial if you recall the derivation right in the derivation this means t final minus t initial so therefore if you will choose B as the uh initial point B as the initial point here you should put TC minus TB. So then so there's some subtle things. So that's why I chose A as an initial then TA equals Z. I don't need to worry about modifying this equation. But but like I said derivations are also meaningful because you know that through the derivation you know that we chose T final to be t and t initial to be zero. That that's why we had such an equation here. Right? But so use it carefully. Hey, so you use this equation only if t initial is zero. If t initial is not zero, you should modify this this t part of the equation. Yes. Which method? I mean which method are you assuming no the distance is the same the time is the same velocity the negative of the initial oh if you want to make okay this is actually by your prior experience if you make such an argument that means previously you may have done some calculations already so that you have such a sense but for general audience if they don't have such a sense it's still better to actually do a computation right but but if you so that's a good sense I mean the the sense that you address is a good sense but I I I guess it was you got this probably from a prior calculation right no or from real like experience okay so yeah that will be good yeah I mean that this sense I mean your argument is right if there's no air resistance no air resistance since the problem is symmetric so from TA to TB the time interval will be equal to the time interval from TB to TC right then you can argue that the directional velocity the VA and VC they just uh they are they are of the same magnitude. Yeah. You if you have such a sense it's also great. Yeah. Yes. But but still we we need some verification and we use the kinematic equations to do the verification. Okay. So good. So now let's move on to the next. So now so so you have great questions. Okay. It's always great to think. Now now let's look at the part D of question. Find the velocity and position of the stone at t= 5 seconds. Right? So this is purely mathematical calculation. So they assume a point d point d is somewhere here. Hey point d is somewhere here during the course of motion and a point d if they give you the time td they want you to obtain the velocity of the stone and the position of the stone. So this is just straightforward calculation. So I don't plan to write on the Okay. So let me I can still write on the whiteboard again but just uh very briefly part D part D of a question. So the textbook they just gave you such an example here. That means this example is meaningful but the rest you can practice the homework problem. So up to this point in principle we have already finished everything for chapter two already right chapter two we will end end the discussion at here but but now let's finish part part D part D of a question so part D of the question like the solution is very similar to the textbook solution because they time is given explicitly so to get the velocity so in part D of the question you assume a to be initial and then b to be final. So again this choice is not unique but like I said we chose a to be initial point so that we don't need to worry about to modify the time this time right this time if ta equals zero then you use this equation this t is just td right so so then you have v d equals uh va minus g * t something like this right and then your vd should be equal to uh 20 m/s then minus 9.8 m/s squared then times the time which is 5 seconds here you have three significant figures so you just omit omitted some some uh but but anyway so now now after you have done the computation you will see this value here right the value is about - 29 uh meter per second here why is it negative now you can see the structure right this 20 here but if your minus gtd is I mean if that size if that this magnitude is greater than VA then your VD will become negative right so that's it so so that means this V D will be ne negative means it will point just vertically downward now if you will compare compare VC to VD which one is greater let's let's recall the numerical values right recall the numerical values here VC we have obtained it early on right was - 20 m/s And what is at some location D here? What is V D? Okay. So think about this whether it's great bigger or smaller. So think about this. So this V D here. So somebody addressed some good argument here. V D here is negative uh 29 m/s. Now let's compare VC to VD and argue that which is bigger, which is smaller. So first we need to think about why why are these two both negative? Why are these two both negative? They are so let me tell you why they are negative. They are negative just because we assume positive yaxis to be up and those VC and VD they are both pointing in into the negative y direction. So that's why they are negative. So now let's compare consider your real life experience. Okay. when you drop this ball like when during the descending process we know that here you have VC is 20 but negative -20 just mean that it's uh pointed downward right and now VD if you try to compare the magnitude of V D to magnitude of VC which one is greater in magnitude V D is greater in magnitude so be careful about this okay so that means this ball is actually speeding up during the descending process because acceleration is down right and VC and VD they are both down. So therefore if you consider the speed of the ball or stone is actually speeding up from C to D. But now when you look at the negative value you will actually become more negative right from VC to V D it appears to be more negative. So be careful about the language here. Right? When you say that consider the speed from C to D. This stone speeds up from C to D. But when you consider simply compare VC to V D which one is greater in their value. Actually actually VC looks less negative but VD looks more negative. So if you simply compare the value of velocities this VC appear to be larger in value than V D right because it's less negative more negative so that's why sometimes we need to use speed sometimes we need to use velocity so be careful about the language here do do you see the meaning here yeah don't get confused so in one of the previous lectures in the first week I asked you to com compare the the velocities two velocities and two speeds. Right? Here you can do the same. Compared to two velocities, VC appear to be less negative, looks greater than VD. However, if you consider real life experience, VD is larger in magnitude. That makes more sense to our uh real life experience, right? The language of speed makes more sense because the language of speed is independent of the coordinate that you choose, right? Speed is just instantaneous speed. It's just the absolute value of the instantaneous velocity, right? Just take absolute value compare the speeds. But but anyway, but here you should keep the negative sign, right? Because they are asking about velocity - 29. And then then in the end you can compute the position, right? So position uh position you just follow the calculation position and so the equation you use to compute the position uh will be like a yd right should be equal to y a and then pl uh plus the v okay plus the va times times the time td then minus 12 the one half the g n times td² right so like I said you use one equation and that will be this equation here and once you have used this equation and you can substitute it in the numerical values here I don't have enough space so I drag it down here so in the end you will get that value of yd here yd equals -22.5 m right so the notation I use is slightly different I use negative g right but here I just say that but they use a y but because a y equals negative g we have such a relation here right so now yd is negative what does that mean so it's not negative it means negative below negative y means that it's below the initial location right but also we can also pay attention that the height of the building is 50 m right then some question they may also ask you what's the what's the location of this point d above the ground level. They can also ask you this, right? So, pay attention that the YD you have obtained is relative to the roof. Right? Now, YD is negative because it's below the roof. It's at a distance of 22.5 m below the roof, right? But they can also ask you what is the location YD, right? What is the location of this point D above the ground? They can also ask such a question here, right? But they they didn't ask such a question here uh in in this problem here. Right? So basically we have finished uh section A already. Now let me briefly comment about section 9 and section 9 is not relevant to our test but some of you may be interesting in reading it because section 9 they they will try to derive the kinematic equations by using calculus. Right? The reason why we don't talk about section 9 is because many of you are still taking calculus too. Right? But this idea about integration is actually relevant to the uh calculus 2 actually. So that's why we skip this session. But actually it's just derivation. It's nothing nothing deeper than the derivation. And if you read this section, you can see the derivation of three kinematic equations. So but that's it. Hey, but this again this section nine we don't cover. We end at section eight. So keep working on the homework. Hey, and so thank you all and hope you all have a great day. See you again next time.