 Hey everybody, welcome to Tutor Terrific. This is the final lesson, lesson five of my chapter three of my physics course We're going to be doing part two of all the projectile motion problem solving techniques We're going to do a problem with many parts and then a final problem And we're definitely going to use all of the formulas today including this guy the range formula will he'll come back and After this lesson, you should be ready to do any introductory college level projectile motion problems or high school rigor rigorous physics course problems in this chapter on two-dimensional projectile motion. Let's begin All right, we're gonna start with these techniques that we've learned so far. I went over this in the end of the last lesson Let's make sure we remember it We're going to determine which object or objects are involved and we're going to draw that picture if there is not a Picture we need to draw one. Okay, next we're going to see sometimes you're going to see this today Sometimes after you draw the picture as easy conceptual answers to the problem Exists and you don't have to do any math that is possible. Also If that's not what happens, we're going to go on to next ask if the range formula is needed or if it applies and That's only if the initial and final vertical positions are the same and you are asked for a quantity in the range formula And you know all the others You can sort of shortcut this list and stop right there And solve for the unknown quantity using the range formula if the range formula doesn't apply We're gonna have to use one of those five 2d kinematic equations two horizontal and or three vertical Equations and you're gonna have to set up your two sets of given and wanted tables one for each component Then you're gonna have to while you do this set a positive and negative direction for the two components horizontal and vertical and I want you to also for the positions set a reference position Where is the zero for horizontal movement and where is the zero for vertical movement where they most help you and you? Usually have the freedom to do so Then you're gonna pick a roadmap. You're gonna figure out which equation Do I need to use to solve for the exact quantity asked for in this problem or this part of the problem and you You're gonna need to find it to move forward So you might have to do some investigative work or use more than one equation to get to your desired result And that's how we do these problems. Okay, just a little review on that and let's get to a problem This is a five-part problem about a kicked football Okay, I'm gonna show you everything you can be asked when it comes to your Kinematic two-dimensional equations and the range formula so first this football is kicked at a 37 degree angle with Velocity of 20 meters per second We have not seen one of these problems where the initial velocity is at an angle And we need to really dive into this because these are the most complicated type of problems. You're gonna see so in this first section Part a we're gonna find the maximum height that the football reaches the maximum height. Okay, the picture is here It's very clear the velocity vectors are shown. That's very nice the initial velocity Vector is shown and the angle is there It's really nice and it were asked for the height the maximum height that the football reaches So can we use the range formula here? While the range formula could apply to this situation We're not being asked for the range or any quantity involved in the range So no we will not use the range formula here We're gonna have to create our set of given unwanted tables What I'd like to do for this problem is I'd like to sense the football initially moves upwards I'd like to set positive as upwards and I would also like to set going to the right as positive because the football in the picture moves to the right And so positive is upwards and to the right Is positive as well. So let's continue By resolving the velocity into components We have to do this because we're given a quantity. That's not Horizontal or vertical it's a combination of both and the initial velocity specifically We will need to resolve any quantities like that into their components before we can complete the tables And so in order to resolve this v not into components We'll just multiply it by the cosine of 37 for vx not Like I've done here in this first row you get 16 approximately meters per second And then you multiply that initial velocity by sine 37 degrees to get the vertical component Which is smaller because 37 degrees is less than 45 And so you get 12.0 meters per second for the initial vertical velocity Okay, and it's positive because I chose positive upwards. So now we're ready to create our tables We are given the initial x and y velocities Also, I'm going to set the initial vertical and Horizontal positions to be zero because in this problem when it asks for the maximum height only a difference matters So I'll set the initial to zero so that the difference is just that value up there that we're going to eventually find So x not on y not are zero also g is negative The acceleration due to gravity is negative nine point eight meters per second squared since it points down okay, and in addition to that the final Horizontal velocity should be the same as the initial because there's no acceleration in projectile motion in the horizontal direction So that's why vx and vx not are both identical and always will be for your horizontal component of projectile motion And then over here we see our given y values Notice that I also added v y equals zero meters per second. Excuse me. I fix that there you go Zero meters per second y because at the top think about it We are turning around when it comes to vertical motion at the very top That split instant in time. We're at the top. We don't have any vertical velocity. We have zero vertical velocity That's a very important understanding That the problem picture makes clear you have to understand that when you're at the maximum height your v y final is zero Okay, now what are our wanted values? Well, I don't know how far it moves Horizontally, I don't know the maximum height which would be the final y value, and I don't know how long it took to get there So these are my unknowns notice how I straddle the t between the two lists because time Exists in both lists because it's in both equations. That's the same t Okay, we've got our table Now that we have our table, let's get our roadmap How can we find the maximum height which is y final? Okay, it's the Final position if you set the initial position to zero it'll just measure that displacement vertically speaking Great, how are we going to get that knowing everything over here? Well, if we look at over here, we don't have any time, okay? And it's a vertical quantity we're asked for so that means we're going to have to use a vertical equation That does not have time in it, which would be the bottom one. Let's just check. Do I have v y? Do I have v y not do I have g do I have? y not yes I have all four of those and why is the only thing I have to solve for So we're going to solve this equation for the maximum height y and y not is zero in this situation Because I've set it to be such as I've said before So when you set this up, we're going to solve for y algebraically first By first subtracting v y not squared to the other side Okay And then dividing by 2g now you might be alarmed saying hey, hey, hey, hey y is 9.8 meters per second squared positive I thought you said g was negative 9.8 meters per second squared notice what they've done here algebraically In the original equation it would be v y squared minus v y not squared notice how they switch those around So they've factored out a negative To in order to switch those around and they've divided it and cancelled it with the negative In the g value and so everything's fine. Nothing's nefarious has been done We just used a little algebraic manipulation to get rid of some of the negatives So this is equivalent to zero minus 12 meters per second squared over two times negative 9.8 meters per second squared So it works fine when you compute this you get 7.35 meters two sig figs Uh, actually three sig figs based on the initial calculations and the fact that I'm doing multiplication So that's the maximum height 7.35 meters with a 37 degree kick That's pretty reasonable with an initial 20 meters per second velocity And that's because it's such a low angle. Okay, it's not going to get that high What you're really concerned with is distance and we know 45 degrees is going to give us the most range But um, this kick wasn't 45 degrees. It was eight degrees less. All right, so that's the maximum height of the football Now let's move on to part b Part b asks for the time of travel before the football hits the ground Okay, so it's asking us for time clearly Would we use the range formula for that? No The range formula has no time information in it. So it's not going to help us Find the time. So we're going to need to Create our two sets of given unwanted tables. We just have to modify the set we had before Okay, we're still going to consider positive upwards. We're still going to consider to the right positive No, the last picture on the last slide But this time I know a few more things This is the entire trip now before the football hits the ground means This is the whole entire trajectory from one side all the way to the other And so our final y is not the maximum height now. We have to alter it and make it Zero the same as the initial because it's the entire trip from the ground to the ground Okay, so now y final is over here also Remember the symmetry I showed you in chapter three unit three Excuse me unit two and I revisited in unit three if the initial vertical velocity is 12 And it's horizontal level situation The final vertical velocity will be the opposite in direction But equal in magnitude making it negative 12 meters per second. So that's how I've altered the given sets The um wanted sets are no different except my y has moved over Final y I still don't know how far it moved horizontally Which would be the range and we could have used that if we were asked for that in this problem what we worked and Time and we still don't know the time but this time we're asked for it. Okay, so the road map is going to involve finding the time So which equation can I use to find the time given what I know? Well I'm going to use the second one. I know you can use the first one But I'm going to use the second one to illustrate something to you Okay, just in case you didn't know about the symmetry The negative 12 meters per second you can use the second one which doesn't involve the final velocity Now the second one looks a little scary I know the initial and final positions are zero so I could plug in zeros for those two numbers But vy not is not zero. It's 12 And so this middle term t to the first power term is there to stay and of course the final one is as well because of the Fact that g is negative 9.8. Yes, the negative was pulled out By the textbook that did this problem So now I've got a true quadratic equation when it comes to t and it's equal to zero But the constant term is zero. So remember from algebra You could factor out one of the t's from these two terms giving you this This is great because this allows you to solve for t two separate ways using the zero product property One of the results gives you t equals zero now. That's Ridiculous because it's not going to take zero seconds for the football to hit the ground This solution refers to the initial position of the football on the ground. Of course, it would take no seconds For the football to hit the ground if it's already on the ground The real solution we want is from this linear factor here one half 9.8 t minus 12 and so if you Set that equal to zero Subtract 12 to the other side then divide by one half which same as multiplying by two Divide by 9.8, which sends it downstairs. You get this t equals 2.45 seconds. Awesome So do not be afraid when uh your middle term in your second vertical equation is not zero You can still factor it even if the first one isn't zero You can always use the quadratic formula negative b plus or minus the square root of b squared minus four ac All over to a that's the solve I used to remember that you can always use that Uh to solve the second equation for t So 2.45 seconds is the time for the entire trip, which is pretty short But um if that ball is going that fast that makes sense so um What I want to tell you something this book didn't do is it plugged in these numbers really early I would not recommend doing this. Okay. It's very easy to get lost in the numbers I would have gone back to here set the zeros to be zero and then solve for t before plugging anything in It's really generally good practice in my opinion to do that because you don't get as lost in the numbers And I've seen people get lost in the numbers and forget what they're doing or what they're solving for I would have waited till this point to t was isolated to plug in numbers personally Okay, that's part b. This is a long one Things get easier now How far away it hits the ground? Okay, so if you think back to the last problem we saw for t Only thing left to solve was for x final. Look what the question part c asked for how far away does it hit the ground? Well, that's x final And since this football is a level horizontal Range problem because the football starts the ground and ends at the ground You can use the range formula to find this final x All right, so we are going to try it We're going to choose positive downwards this time, which is generally good practice And choose right as positive because there's no reason not to This will allow g to be positive Okay, we want g to be positive in here so that we don't get a negative r And it's also going to mess up our angle as well So we're going to choose positive downwards when we use the range formula generally speaking The two quantities we know are the angle theta and the initial velocity being up 20 meters per second and 37 degrees. We're finally using just the bare bones and initial quantities given to us Now we plug them into the formula like this and g is 9.8 meters per second squared as we always know Positive because positives downward We have to multiply the angle we plug it by 2. Okay, you know, I forget that so we're really finding the sign of 74 degrees I didn't get too much into the solution of how this is derived, but the two theta is absolutely necessary So when you compute this you get the following 39.2 meters. Okay Let's say you didn't use this. Let's say you didn't think of the range formula There's actually another way to do it When you look at these equations that we found every possible Uh variable except for x final That means I could easily use this first equation the second horizontal equation here. Excuse me to really find this value Okay, so you would just modify your given sets of given and wanted tables so that the time was moved from here To here you'd see that x equals question mark was all that was left Meaning you could use this equation right here. That's the roadmap You could plug in vx not and x not is zero and you could plug in the t value you got 2.5 four five seconds from the last problem And you will get 39.2 Also, so that's probably an easier way to do it But of course some people are going to go to the range film because they're scared of this table of five equations And the range formula seems like an outlier that's easily used And that's fine. You get the same answer both ways your choice Okay, the last two parts What is the velocity vector at the maximum height for the football? Okay, this is a purely conceptual question You need to always look at the drawings and ask yourself. Is there a conceptual Answer to the question Look at this. Look at this picture. Okay. This is the original image. I showed you look at the maximum height velocity vector It's purely horizontal guys. It's purely horizontal So what does that mean? That means that the Velocity vectors horizontal value is the entire velocity vector But what's even better is that the horizontal velocity is constant in projectile motion So it's initial value of vx not is equal to this answer that we're trying to find Okay, we already found This particular Value it was 16 meters per second and so the total velocity vector at the top Pure horizontal has to be 16 meters per second as well So all the work was already done. That's truly a whole conceptual question And the last one. What's the acceleration vector at the maximum height? Okay, so again Is there a way To conceptually answer this question. This one is honestly just Trying to trip you up and make you think oh I have to do a bunch of work. Oh my gosh. Is it very Does it is a the acceleration different at the top than anywhere else? No Not for our um nice beautiful world in which there's no air resistance the acceleration vector is constant everywhere And I know um Those of you who know this stuff very well and you've studied Newton's law of gravitation know that it's slightly different when you're farther away from earth's surface But this is we're talking seven meters, okay, this is not uh, this is negligible. So we're going to uh, Like we said, we're going to treat the acceleration vector as constant in these problems Near the earth's surface. So the acceleration vector at the maximum height is the same everywhere Not just at the maximum height. It's the same at the ground It's the same halfway up. It's the same halfway back down. It's 9.8 meters per second squared Okay, and it points downward. So the full question's answer would be 9.8 meters per second squared downward So that was a pure conceptual question. So it seems like these parts of this problem kind of got easier as we went along Okay, we finished that long and drawn out problem. Sometimes you have those on your tests or on your books This particular book the geoncaly physics six addiction Sixth edition, excuse me Was quite full of those types of many multiple part problems. Okay one more different problem Suppose one of napoleon's cannons had a muzzle velocity of 60 meters per second What angle should it have been aimed to strike a target 302 meters away? Okay, all right So We had to draw a good picture now this would involve some cannon being shooting something And then it landing on the ground somewhere later. Did you see anything about a uh cliff or anything? No I saw that y final equals y not which means the initial and final positions are the same And it's asking for an angle to aim the cannon so that its range is 302 meters So when I ask the question after the picture is drawn in my head and you've drawn on your papers Would you use the range formula? Oh, yeah, baby That's the only one that has an angle in it And so that was a good hint But I needed to check that my initial final vertical positions were the same so I can use the range formula. Yes I'm going to choose positive downward And uh to the left positive. It really doesn't matter here. Um, let's if I drew the picture such that the cannon launched the Cannon ball this way that would make sense And uh, so we're going to use that range formula We've got to find the angle this time Our range is 302 meters given to us our initial velocity is 60 meters per second. We've got to find that theta So first things first Please uh solve for sine 2 theta solve for the sine of the double angle Um by multiplying both sides by g and dividing both sides by v not squares We ended up with this On one side Now in order to move forward in order to get to the angle. I have to do the inverse operation to sign I have to do the inverse sign To move forward. So uh the inverse sign Will be uh of 0.821 repeating when I create this and I plug in 302 9.8 and 60 squared Um, and I compute that fraction I get 0.821 repeating Then I take the inverse sign Of that This will give me the angle 2 theta which is inside The uh sign right now, okay And when I get that I get 55.297 make sure your calculators and physics class are in degree mode There isn't really going to be a time when you need them to be in radian mode That's more for your pre-cal trade classes and your calculus classes So when you go to physics class radian mode is the mode to be in Then we need to divide that 55.297 divided by 2 to get theta not 2 theta And you get 27.6 degrees with three significant figures That's how many I have in the initial quantities So That's one angle. There's actually two answers. Okay, if you remember my conceptual drawing Um, this model I had you saw that the complement of each angle Launch angle also gave you the same range. So to be fair, there's two angles That will give us the same 302 meter range. The other would be 90 minus our initial answer 27.6 the complement That's just 62.4 degrees both of these angles give us a 302 meters range With a 60 meter per second launch velocity, okay So remember that there are two angles because the complement of the angle we got delivers the same range All right guys. Thank you so much for watching This video is now finished. You have finished the entire unit on two-dimensional kinematics and projectile motion Thanks for sticking through it. Next chapter is going to be on forces for now. This is falconator signing out