Ch 04: 2D Motion (Projectile Motion)WorksheetSee all chapters
All Chapters
Ch 01: Intro to Physics; Units
Ch 02: 1D Motion / Kinematics
Ch 03: Vectors
Ch 04: 2D Motion (Projectile Motion)
Ch 05: Intro to Forces (Dynamics)
Ch 06: Friction, Inclines, Systems
Ch 07: Centripetal Forces & Gravitation
Ch 08: Work & Energy
Ch 09: Conservation of Energy
Ch 10: Momentum & Impulse
Ch 11: Rotational Kinematics
Ch 12: Rotational Inertia & Energy
Ch 13: Torque & Rotational Dynamics
Ch 14: Rotational Equilibrium
Ch 15: Angular Momentum
Ch 16: Periodic Motion
Ch 17: Waves & Sound
Ch 18: Fluid Mechanics
Ch 19: Heat and Temperature
Ch 20: Kinetic Theory of Ideal Gasses
Ch 21: The First Law of Thermodynamics
Ch 22: The Second Law of Thermodynamics
Ch 23: Electric Force & Field; Gauss' Law
Ch 24: Electric Potential
Ch 25: Capacitors & Dielectrics
Ch 26: Resistors & DC Circuits
Ch 27: Magnetic Fields and Forces
Ch 28: Sources of Magnetic Field
Ch 29: Induction and Inductance
Ch 30: Alternating Current
Ch 31: Electromagnetic Waves
Ch 32: Geometric Optics
Ch 33: Wave Optics
Ch 35: Special Relativity
Ch 36: Particle-Wave Duality
Ch 37: Atomic Structure
Ch 38: Nuclear Physics
Ch 39: Quantum Mechanics

Concept #1: Intro to Projectile Motion


Hey guys we're now gonna talk about projectile motion which is a really big topic in physics and a lot of students struggle with it let's check it up. All right. So projectile motion happens when you launch an object and it moves freely in two dimensional space. It's different from one dimensional motion. If you throw something up it goes straight up and it comes back straight down. We can call that just vertical motion. But if I toss something at an angle it's going to move something like this. And that's because as it's moving it's being pulled down by gravity. So that's why you have that Parabola path. And this is one dimensional motion and this is 2-dimensional motion in the x axis and y axis at the same time. So we call this projectile motion. Ok, so projectile motions any kind of motion when you throw something and it moves in two dimensions. One of the challenges of this is that there's all kinds of different variations in all kinds of different ways in which you can solve these problems so we're going to do is I'm going to show you the main types and also try to develop a systematic way give you a systematic way for solving these. All right. So remember that the first step in solving any two dimensional physics problem whether it's motion projectile motion forces is to actually decompose the problem into x and y. Okay so we gonna decompose the problem into an X and Y parts. So let me show you that real quick here. So if I were to launch something here I'm going to call this point a. And I'd throw with the initial velocity Let's call this as Va at a certain angle and we'll call this theta A. This object is going to have this path here given by the dotted line. Right. And so the if you kind of continue here this point here is important that's the maximum height be in this point is important over here as well because that's when you hit the floor C . So what I'm going to do is I'm going to imagine this one motion one two dimensional motion as two one dimensional motions. So I'm going to imagine this one two dimensional motion as two one dimensional motion. One in the X-axis and one on the y axis. So imagine that I'm going from A to B and back to C but I can only move in the x axis if you want to go from A to B and then to C but you can only move in the x axis. It would look like this. Right. Imagine these tracks and you can only move along the x axis that's what it would look like. And B is somewhere here. Now if you're doing this in the y axis if you're going from A and then to b and then to C but you can only move along the y axis. It would actually look a little bit like this. I'm going to move it over here. This is your point A and you would be going up. Hitting B and coming back down. So this would be A A C there's is a B there. A, B and C. So the idea is that projectile motion decomposes into motion in the X-axis one dimensional motion the x axis and going up and down one dimensional motion the y axis. And if you treat as, it has two dimensional motions it's much simpler. So that's what we're going to do now when you do this you do have to decompose your initial velocity. Right. Because remember vectors have to be decomposed. I can't do any kind of math with a vector like this. I have to decompose it into x and y. So I would split this into Vax and Vay now remember Vax if the angle is with the x axis which it is is Va cosine of theta a and Vay ups that's Vax , Vay is Va Sine of theta a. So that's the first thing you do , you split it into X and Y and you can kind of imagine this motion going in the x and the y axis as you get better with this you don't have to draw the two separate ones we're going to start this way.

So when we're decomposing it says here let's continue when you decomposing initial velocity which here I called Va but it could be Vo or Vinitial Dox goes with the x motion and Voy goes with the y motion. What do I mean by that. Well whatever this number here whatever number you get for Vax that's the number that you're going to use in the x axis whatever number you get for the Vay. That's the number that you're going to use in the y axis and this Va you actually never use it again. Right. Once you decompose that vector you're not gonna use Va anymore. So again, I'm splitting this one problem into two problems with the blue lines right. X and Y. And these are the numbers are going to use this here and Vay over here. So for example here this velocity is my Vay. And then this velocity over here is my Vby. And this velocity over here is my Vcy. OK. All right. So projectile motion as I've shown you is basically vertical motion. long the y axis with constant velocity. Horizontal motion right in the. What's special about the x axis is that there is no acceleration. And obviously, the acceleration in the y axis will be gravity, we will talk more about that than the second. These two motions you should know are independent and synchronized. Synchronized means they happen at the same time. So if you think of splitting this into two motions when you are in the X-axis at point B is the same time when you are at point B on the y axis as well. OK. And independent means that what's happening the x axis has nothing to do with what's happening in the y axis, the only thing that connects them is time. So you can look at them as completely independent however synchronized motions. OK. And then the last thing I want to talk about is that remember that if two points for example A and C are at the same y value if they're at the same height like these two points here are right. A and C are at the same height. Then they are what we call symmetric. The motion between those two points is symmetric. And if you remember what that means is that the time to fall will be the same as the time to go up. Those who time are the same and the velocity coming down to that point will be the same as the velocity going up except that they have opposite signs because they have opposite directions. So symmetry is important you should know that. And then one more point actually is that at any point the object's velocity vector is tangent to its path. So what does that mean. If you want to draw a velocity vector and you know what is the velocity vector look like right here. Well it's tangent to the path. So it looks like this right of the path like that. At point B tangent to point B. Looks like this. So this is what Vb looks like. And Vc looks like this. Write all it's change into the path. If you pick a random point over here. Its velocity vector looks something like this. OK. So this is a V at a random point V at some other random point A, B and C are the important points here. So let's talk a little bit more about some other conventions we're going to use and these problems I'm going to say that going up into the right is the direction of positive. OK. In the X-axis we already mentioned briefly that the acceleration is zero because the acceleration is zero I know I don't have three equations I actually only have one equation right the equations simplified to just one. I am gonna actually write that equation right here which is that Delta I'm sorry that V average. V average which is a constant velocity is delta x over delta-T. But if I move this around. I get delta x equals the XT. And this is the x axis equation for projectile motion then I'm going to use over and over again. Again it comes from here. But I'm just going to use it in this second format here.

OK. Because the acceleration in the x axis is zero. This means the velocity in the x axis never changes. So I can see the velocity at point A is the same as the velocity of point B and the same is the velocity of points but the velocity in the x axis. So that's a really important point. Don't forget that so Vax is Vbx which equals Vcx. So instead of calling it Vax, Vbx, Vcx I'm just going to call it Vx which is why I put it just a Vx. So the x velocity never changes because the acceleration is zero. On the y axis the acceleration is gravity which is going down. And since we're saying that going up is positive the acceleration the y axis would be negative gravity. Remember gravity is 9.8, positive but the acceleration the y axis is negative gravity so it will be negative 9.8. And because the acceleration is not zero. I have the three, the typical three to four equations of motion right. Three sometimes four equations depending on whether you're professor let's use the fourth equation we have talked about. OK. So that's the basic idea here as you're going along the x axis the acceleration is zero. And as we're moving along the y axis the acceleration is negative G which is going down. What else do we know here. This Vay comes from here. The velocity Vby is zero. Just like how it wasn't vertical motion if you throw something up it reaches the maximum height the velocity is zero comes back down. And in this particular case this is a symmetric launch. Right. So there's lots of types of different projectile motion problems. This one I refer to is a symmetrical launch, symmetric launch is where your final height equals, I should actually to say Yfinal equals your Yinitial. Ok. So you return to the ground you return to the original height. And that's a special type of projectile motion problem. It's one of the simple ones. And what's nice about it is that thereÕs some symmetry here. So Vay I can get it from here. And once I know it this Vcy is the same as Vay because of symmetry I also know that this time to go up over here is the same as the time to go down so I can say Time to go down from B to C is the same as the time to go up. And those are because of symmetry. OK. So enough of that lets do a problem. We Really do lots of problems to practice this stuff. So one thing that to mention right before we start is that every time you get asked the question usually you get asked two-three different things about motion. You're going to sort of follow these steps. For every question that you're asked you have to first figure out. What x is the variables in. So for example if I ask you give me Delta X. obviously that's in the x axis so I'm going to use the x equation but if I ask you what is Vfinal in the y axis. Obviously this is on the y axis. So I'm going to use equations from the y axis once you know what axis to use. You have to figure out which interval to use. Right. So now we'll kind of do that in the example then to pick one of the three equations or one of the four equations and figure out what to do. All right so let's get started to do this real quick. So an object is launched from the ground with 50 meters per second and 30 degrees. This is positive 37 degrees. So it's above the x axis and it later returns to the ground later returns to the ground means that this is a symmetric launch. The name is not that important. What's actually important here is that you know that you're going to be able to use symmetry between the very beginning to the very end. So I want to know the maximum height. All right. So maximum height you need to figure out what variable this is, maximum height is the height from the bottom all the way up. So it's delta y. Right. If you think about your five variables of motion which are. Which are Vinitial, Vfinal, a, delta x and delta t. Right. Or if you're on the y axis this will be Delta y. So maximum height is your Delta y now it's Delta Why is the change in position from where to where from a to b. Right. Look into diagram. So what I want is Delta y from a to b then I want to know how long it takes to get there. So I want delta t from a to b and then I want to know what is the total time of flight. I will talk more about this later but the total can be thought of as time up plus time down. In horizontal displacement. It's also referred to as range is simply your delta x. Okay so let's do this one at a time. All right. So maximum height Delta y from A to B. Let's look at the steps here again I want to give you a systematic way of solving these. Look at the axes. Is this going to be in the x axis or the y axis, this is obviously going to be in the y axis. That's the first step. The second step is to pick an interval. Here there's not much of a choice. I want to know the maximum height from a to be. So it's going to be from a to b. Nothing more that you can do there. And then you just have to pick an equation. OK, Now before I pick an equation I'm going to draw this motion just from a to b only from a to b and on the y axis only. So what I'm doing is I'm getting this entire problem of a to b to c in 2-dimensions and turning into a one dimensional problem on the y axis only from a to b only. Ok. So again going through this very slowly because I want to build a systematic way of doing this. All right. So this is I'm going from point A to B along the y axis only. What kind of information do I have. I have to find the velocity of A. But I'm on the y axis so I have to find Vay. And then there's Vby. So let me go ahead and decompose to do that to find these velocities, Vax is Va cosine of theta a. So it is 50 cosine of 37 degrees. You put this in the calculator you get 40 meters per second and Vay is 50 sine of 37. Put this in the calculator you get 30 meters per second. So that means this velocity here is 30. OK. It's positive because it's going up. And Vby is zero because the velocity in the y axis at the highest point is always zero. So I already know two variables. What else do I know. In between these points there is an acceleration there's a change in Y and there's a change in the t, time elapsed so the acceleration we're talking about the y axis the acceleration in the y axis is always gravity going down. So it's going to be negative because going up is positive. We're gonna always do that and these problems. Right. So this is negative 9.8 Delta Y is what I'm looking for. In delta t we're looking for a little bit later but for now we don't care about delta t. Right. While I'm solving part (a) delta t remember is my ignored variable and I can use that to figure out which equation to use. So in fact the equation we uses the second equation and I can write Delta y equals the initial t plus half of At squared. Sorry. Actually that's not the second equation thatÕs the third question. Vfinal equals Vinitial squared plus 2a delta x if I want to Change this for the y axis then it's just a matter of, I can rewrite them or I could have written it already in the y axis. But basically, final velocity is Vby squared initial velocities Vay Square plus true negative 9.8. And then my.

Delta y which is what we're looking for. Ok, if you plug in your numbers the final velocity here is. Zero. The initial velocity is 30 Square. This is minus 19.6 Delta y so Delta y is 900 divided by 19.6. If you carefully move around the numbers of negatives you'll see that the negatives will cancel and I get 45.9 meters. So, 45.9 meters is the answer for your Delta y from a to b. That's part (a). Part (b) asks for. Part (b) asks for delta t form A to B. And delta t from a to b is simply this guy right here. OK. So that's what we're looking for now. Do I have enough information. I do actually know four things now because now I know that t is. I'm sorry Delta y is 45.9. So I'm going to be able to use not just one equation but two equations because I know four instead of three variables. OK. So. Let's see I can use the first equation Vfinal equals Vinitial plus At Vfinal were in the y axis Vfinal is Vby. Vinitial is Vay minus Gt. Right. And that's because if you want to do this in a different way you could do it like this acceleration is negative G. All right. Now let's plug in numbers. And final velocity in the y axis 0. The initial was 30 right there positive minus 9.8t. And if you solve for t you get the signs move it around carefully. The negatives will cancel. And you get 3.06 seconds. All right. So I got that. Now we're looking for next part we're looking for the total time of flight. Again I said total time of flight could be time up plus time down. But we know that the time up equals the time down because it's symmetry. So I can say that time Total in this situation is simply twice the time up. So it's just this number we just got times two. So 6.12 seconds. I'm able to do this because of symmetry. Symmetry. Part (d) asks us for the range delta x ok. Now everything we've calculated so far we've calculated it from A to B right in the y axis but delta x is obviously not on the y axis its in the x axis. So. And if I want Delta x I want Delta x for the entire horizontal displacement. So I want the x axis from A to C , could draw a little diagram here. Doesn't hurt. A pass B, C here and the third step. So I got the interval done the third step is to pick an equation and the nice thing about the x axis is that there is no equation choice there's only one equation. Remember I told you that the x axis equation will be used as just this one every time. Which comes from the definition of velocity v equals Delta x of a delta t . So we're looking for Delta x I just have to have Vx and T and I actually have both of those. Delta x Vx. Look around Vx is. 40 Am time now. When you're plugging in time you have to make sure you plug in the correct time. This is from A to C so I can write the time from A to C right there. So the delta x from A to C and I need to use the time frame to see the entire time 6.12. And if you multiply this on the calculator you get 245 meters. OK. And that's the end of this question. Hopefully this is a soft introduction into the into projectile motion makes sense. I have one last point to make here, symmetric launches started at point go all the way back to that same height attain the maximum range. This is just something for you know when your angle is 45 degrees. Right there's an equation that right at the equation at a drive to show you that. But I'm going to say that it's the theta is 45 for maximum range and the other point is that complimentary angles for example 30 and 60. They will attain the same range. So if you shoot something with let's say 30 degrees and try to make a little bit better. 30 degrees looks like this with the same initial velocity just different angles. You will get the same range. Some like that. Right. So this is 60 and this is 30. You get the same. As long as theta one plus theta two equals 90. That's what it means to be complimentary. So this could have been like 20 and 70, they will attain the same range. All right. So, thatÕs it for that one.

Example #1: Horizontal Launch


Hey guys we're now gonna talk about a type of projectile motion problem called horizontal launch or zero angle launch. And this happens basically when you throw something not in an angle like this but in a straight horizontal line like this. So let's check it out. So horizontal launch aka zero launch is instead of throwing like this. We're gonna launch something like this. So it's gonna move like this hit the ground and all that. What's special about this is since I'm throwing in a horizontal axis the initial velocity on the y axis will be zero. That's the big thing to know here. And because that velocity will be zero your equations will be simpler and this becomes basically the simplest and easiest type of projectile motion problems we could have. All right. So another thing you have is that since the initial velocity in the y axis is zero, your initial velocity is composed entirely of your initial velocity in the x axis Vox. But the velocity in the x axis and remember never changes. Ax equals zero. So this is your permanent velocity Vx. So if you shoot something here with 30 that's your Vx and that Vx will be 30 everywhere. OK. I have here two boxes reminding us of the steps of projectile motion problems. Pick the axes the equation I'm sorry axes integral equation and then the four equations of motion we have. So let's do a quick example and see how this works. An object is launched horizontally right it's launched horizontally. This tells us that the initial velocity in the y axis is zero. And it looks kind of like this. It's launched from a 50 meter high cliff or a building or whatever. So it starts here and it's launched the horizontally so it looks like this flattened the X-axis with 30. So this 30 is my initial velocity I can think of this as my initial velocity in the x axis because it is entirely on the x axis or I can think of this as my Vx which never changes. These are all the same. Okay and this is 30 as well. So this guy's going to move like this and hit the ground somewhere over here. I can call this A and B or initial and final. There's only two intervals. So I'm just gonna think of this as initial and final, thereÕs only two points rather one interval. So I want to find a time that it takes to hit the ground so it's just the time to go from initial to final okay, delta t. And part (b) is asking for range which is delta x.

Now what's unique about time is that you can find time in the X-axis or the y axis. So when I'm determining my Axis time could be either one. So what I want to do is I want to first check the x axis and the reason I'm going to do the x axis first is because the x axis is easiest right there's only one equation and I don't have to even worry about picking one in an equation. So Delta X equals Vxt. From looking for Time. I just have to have Delta x and Vx. Here I know Vx its 30 but I don't know delta x. So unfortunately, I'm not going to be able to solve this using the x axis but that's fine. Now we know for sure that it's going to come out of the y axis so time is going to come from the X or the Y. So now we're gonna have to go to the y axis. Let me draw what this looks like at the y axis if you're limited to the y axis and you're going from beginning to end it would look like this right. So I'm gonna draw it like this. And this is my initial velocity on the Y. This is my final velocity on the Y and I know that my initial velocity in the Y is zero. The final velocity I don't know. This has a high of 50 so my Delta Y is 50. Now be careful going up is positive that's our standard. That's the conventional we're using you're falling so your Delta y is going down so it's negative 50 meters. What else do we have here between these two intervals there's a time and that's actually what we're looking for in our fifth variable motion that we're missing here is the acceleration on the y axis which is little g little g's going down going down is negative in this case so little g. So the acceleration is negative g OK. So I need three out of five and I have three out of five. One two three. Vfinal is my ignored variable put a sad face there and to find out delta x I just have to pick one of the equations and I'm going to use the equation that doesn't have the final which in this case the third equation here will do. All right. So equation number three Delta X equals Vot plus half of at square. This is sort of the original version of the equation but we're talking about the y axis. So I'm going to change to some delta Y. And so this becomes the initial velocity the y axis which is actually zero. So this whole thing is gone, plus half the acceleration in the y axis is negative gt square. So I know my Delta Y is negative 50, this equals negative. If you combine this you get a negative four point nine t square. And if I move things around t is the square root of 50 over 4.9, 50 over 4.9 I get 3.2 seconds as my t. And this is part (a), part (b) is asking for Delta X, Delta X is obviously if you go through your steps. Determine the axes is obviously the x axis. Pick an interval. There's only one interval here beginning to end and pick an equation. The only equation for the x axis is delta x equals Vxt. So to find out delta x I only need Vx which I have and C which I now have as well delta x equals Vx, V is 30 and the time it's 3.2. So delta x is 96 meters. All right. Pretty straightforward I think hopefully you agree. And I want you guys to now try a practice problem. Very similar set up horizontal launch just a different set of numbers. So let's give this a shot.

Practice: An object rolls from the top of a hill (shown below) with 20 m/s and takes 4 s to hit the floor. 

(a) Find the object’s range. 

(b) How tall is the hill?

Example #2: Horizontal Launch


all right guys work out one more example here I have an airplane moving horizontally at five hundred meters above the ground with an unknown speed so let me draw this real quick here's the airplane and fire it's moving horizontally which means its velocity is this way with a v that is unknown a crate that's released from the airplane by the way the airplane is at five hundred meters above the ground.The crew a crate that's released from the airplane travels a horizontal distance of four thousand before hitting the ground OK so it moves four thousand in the X. directionfour thousand meters thats our delta X. horizontal distance before hitting the ground in then asks What is the plane's horizontal speed right so remember when you drop something from a moving vehicle the object will borrow the velocity of the vehicle so the vehicles moving the planes moving this way with v the object to have an initial velocity coming out of the plane as V In fact this is the same thing as if you were on top of a building I was five hundred meters tall and you toss something horizontally with the v it's the same exact situation so it would fall this way so the plane the crate leaves the plane with v and it travels horizontally and hits right here, all right so it's thesame exact situation remember when you drop objects from a moving vehicle and it's asking for the plane's horizontal speed which is the same as this V. here they'rethe same and that's actually going to find We're going to find the initial velocity of the crate and in doing this we know the initial velocity of the plane OK Remember this is a horizontal.

The problem here is on the launch problem right you're launches something horizontally or you drop either moves horizontally which means that the initial velocity in the Y.axis is zero and it means that this is you or vx and that V.X. remember never changes so there. That happens what we're looking for so we're looking for the initial velocity which is the V.X. V.X. is the next variable so I go to the only exit question I have which is delta X. equals V.X.. T in as long as I have Delta X. and T. I can find V.X. if you look around you have the delta x that's four thousand but you know have to be so you're stuck you can't find t you have to go now go to the Y. axis Y. because the X. X.only has one equation it didn't work out so the only thing you can do is go to the Y. axis to try to find.To try to find the time so here in the Y. axis we're going to try to find the time that it takes for this thing to fall and to simplify this a little bit of a draw a little Y. axis diagram. I start here initial final this is my initial velocity the Y. axis which is zero because it's a horizontal launch the final velocity the Y. axis which I don't have I am falling five hundred so my delta y is five hundred except it's negative because I'm going down and remembering these problems going up is positive going to the right positive just a convention we are using and delta T. is what we're looking for what else do I have the last variable that I'm missing here is my acceleration which is gravity but negative G. going down OK if you look around notice that I know three out of five things so I can solve for time and I can solve for time using the third variable because V.y is my defined as third variable third equation rather delta x the initial T. plus half of at square if you do this for the Y. axis it's actually Delta Y. which is five hundred negative.

OK the initial velocity is zero and then I have half of acceleration which is negative nine point eight t squared if you move everything out of the way to get that t you get that t equals. Yes. Ten point one OK so just do the algebra there carefully and you get ten point one But that'snot what I was looking for that was just so I can plug this in here OK let's go back to delta x I'm looking for V.X. so V.X. equals delta x over t your delta t and this is going to be four thousand divided by ten point one So the X. equals three ninety six meters.Per second and that's the final answer for part A anyway so a lot of projectile motion problems will work like that you start let's say in the X. axis you get stuck you go to theY. axis and vice versa are you going to have to jump around it you know one of the things that makes these a little challenging the next question asks for the creates finalvelocity final velocity magnitude and direction now remember visually the velocity is tangent to the path so the velocity vector looks like this is the velocity at a randompoint but I want to know the velocity at the final velocity which is going to look like this.

Remember any two dimensional vector right whether the velocity, force whatever is made up of its X. and Y. components so v is made up of the final X. and the final for me to be able to find v or your theta final I'm going to have to know the first one have to know V.X. and v y so the way that you find v and theta is by knowingV.X. and you know why because remember in a situation like this v the magnitude of your vector is for any kind of two dimensional vector is you get that using pythagoras theoram so you have to get its components first and for theta you need the arc tangent of in this case V y over V x so I have to know V Y and v X. So let's get thosefirst OK So to find the final velocity. And the final theta I have to first get V.X. final and V y final Vx. never changes so the three ninety six that I had is my initialvelocity as well as my final velocity and that's because again the acceleration in the X. axis zero so it's always the same in fact that's why we don't even call this v initial X.or v final X.

we just call all of the V.X. a simple V.X. Now I do need to find a V y and it is the velocity after you drop five hundred meters and it's actually right here this guy OK that's what I need so I already know a lot of information about this already know three things actually are you know four things I know this is ten point one so Ican just use one of the variables one of the equations of motion for this diagram for this interval in the first equation the simplest one will work so the final velocity equalsthe initial velocity plus eighty. the initial velocity in the Y. axis is zero so the final velocity is just gravity nine point eight negative times time time is ten point one so I get that the final velocity in the Y. axis is negative ninety nine it makes sense that I got a negative because. This velocity right before it the ground is going down so it should be a negative OK now that I know the why and V.X. I can finish this so really is the square root of V X. three ninety six plus the square of V Y it's negative ninety nine though it's going to come a positive anyway so it doesn't matter in the answer here is four zero eight. Meters per second.And the angle is theta is the arc tangents of Y. over X. arc tangent of negative ninety nine over three ninety six and if you plug this in the calculator make sure characters indegrees not radians. Fourteen degrees is what you get actually you get negative fourteen .why negative fourteen? Remember when you do work to tangent you have to make sure that the the angle is the right so you don't have to fix it I am in the fourth quadrants this vector here is in the fourth quadrant right one two three four in thefourth quarter is one of the two quadrants that works for the Arc tangent function of the arc in your function works without needing any adjustment so negative fourteen is my final answer I don't have to touch it up.And that's it we're done with this one.

Practice: When an object that is launched horizontally hits its target, its velocity has horizontal and vertical components of 80 m/s and 60 m/s, respectively. Find the object’s range.

Practice: In a regulation-sized beer pong table, the ping pong ball is tossed from a horizontal distance of 2.4 meters and 1.0 meter above the top of its target cup. What horizontal speed must you throw the ball with, so it makes the cup?

Example #3: Negative Launch


Hey guys we're now gonna talk about a type of projectile motion where you throw something in a direction that is below the horizontal so sort of down this way. I call this negative launch because it has an initial angle that is negative as that the negative angle. So if you throw something down this way this is your initial velocity. Remember we always want our angle with the x axis so this is the angle that I want in that initial angle's negative so negative launch the vector is at an angle so I have to decompose it into the Vinitial x. And Vinitial y Vinitial x will be positive Vinitial y would be negative because it's going down. And remember this is your Vx. It's always gonna be the same that Vx will never change. OK. There's in these problems you start with a vertical velocity pointing down. We already talked about that and we usually have just one interval So for example here you're going to leave the top here and basically become a projectile and kind of go like this and hit the ground somewhere over here. So just two points one interval. All variables are down. So your initial velocity in the y axis is down. Your final velocity in the y axis is down to your velocity in the y axis always going down. Your delta y is going down because you're falling and your acceleration and projectile motion problems all of them really is always going to be going down with gravity. So I can say that these are all negative because we establish that going up is positive. Now this is a good example where saying that going down is positive might be a good idea because if everything is going down and I say going down as positive instead of having a bunch of negatives all of my variables will be positive, you could do that if you would like but just for the sake of consistency I'm gonna stick with what with what we've been doing and say that going up is positive and we'll just deal with the negatives. OK. So here I have an object that slides off an inclined roof. The English known as 37. It comes off the roof with five meters per second three meters above the ground. So the idea is that if you're sliding off the roof this way when you come off the roof you're gonna have a velocity point in that direction. So this is really the same thing as if you were on top of a three meter tall house or whatever and you threw something down, so instead of it rolling down on its own. You threw something down this way at an angle of 37 below.

With an initial velocity of 5. It's the same thing, but it rolls so the angle of the roof becomes the angle of the initial velocity here just some basic geometry. So this is actually 37. That's my initial theta. And my initial velocity is 5, ok. I do have to decompose this into Vinitial x which again is just my Vx and to do this Vinitial x or simply Vx is Vinitial cosine of theta initial. So it is five cosine of Negative 37. We should expect this to be a positive number. So and if you do this you do get a 4 which is positive will get a positive 4 action. So Vinitial on the y axis is 5 sine of negative 37. And the calculator will give you a negative 3 which if you look in the picture if you look on the little diagram here it makes sense that this velocity is negative because it's pointing down, ok. So Voy is negative 3. Why I could have just written that 3 because the error already indicates that's going down. OK. And then this guy is a 4. Cool. So, I want to know first thing we're looking for is the horizontal distance, horizontal distance is obviously my delta x. LetÕs Look at the steps I'm supposed to pick an axis so obviously delta x is in the x axis. Pick an interval. There's only one interval and pick an equation. The only equation that I have for the x axis is delta x equals Vxt. So there's really nothing to pick. Just knowing that you're supposed to use that. Alright. So I'm looking for delta x. All I need is these two variables and if you look around I have Vx, Vx is four but I don't have t. So I have the x but I don't have t and as always I get stuck in one axis I'm gonna go to the other one. So I'm gonna go now to the y axis. Looking for t, ok and I can do this by drawing what's happening on the y axis. So this is my initial velocity on the y, my final velocity on the Y, the initial velocity on the y is 3. But it's going down so it's negative 3 and the final velocity in the y I don't have that. And I'm looking for delta t. The drop is 3 meters. It's going down so it's negative 3 because I said remember going up is positive and the acceleration is gravity negative G. Because we're going down. This final velocity here will be negative. But I don't have it so this is my unknown variable. Actually I'm sorry this is my ignored variable my unknown what I'm looking for is t, ok. so I can use an equation to figure this out and the first equation will work. Sorry the third equation will work. Delta x equals Vinitial t plus half of at square. Now if you try to do this you're gonna get a quadratic equation right. This is going to give you a quadratic why because we're looking for t and there are two t's. Right. Because you have two unknown t's here. This is going to lead to quadratic equation. I have two options you can embrace the quadratic and solve it or you can try to avoid it if you don't want to use equation 3. Then you're going to have to use equations 1 and 2 and just for the sake of showing you that process I'm actually going to avoid the quadratic and instead use equations 1 and 2. OK. So I'm gonna start off here with the equation number 2 and what that's going allow me to find is it's going allow me to find the final velocity. Which then I can plug into equation one. OK Vfinal equals Vinitial square plus 2a delta x, the final velocity is what I'm looking for. So, Vyfinal square the initial velocity in the y axis is negative 3. It gets squared so it's going to lose the negative.

True the acceleration is negative 9.8 and the drop and the drop is negative 3, ok. So when you do this you get that VY square is, see where they get here you get that Vy. So you're gonna simplify all this Vy would be the square root of some stuff. And then when you carefully do all this algebra you get an 8.23 meters per second. Now you have to be careful here. Remember every time you get a number coming out of a square root like for example square root of nine this is technically plus or minus 3. So this number could be a negative. And that's because three squares a nine and negative three square is a nine as well. So, even though you might think that this is a positive look here your final velocity is expected to be a negative. So you just gonna force a negative in front of this, Ok so your final velocity in the y axis is negative 8.23. Once you do this you can now use the first equation to find the time, so its a work around, work around the quadratic equation. Vfinal Vinitial plus 80, the final velocity is negative 8.23, the initial velocity in the y axis. Remember we're doing all of this just on the y axis is negative 3. The acceleration in the y axis is negative 9.8. And now I have all the pieces to find time. And if you move everything around you get that time is 0,53 seconds, 0.53 seconds however that's still not what I was looking for, this time here was only so I could plug it back in this equation to find delta x. So let's do that finally Delta x equals Vxt, Vx is 4 positive, time is .53. So the answer is 2.12 meters, ok. The answer to part (a) is 2.12 meters. Let's make some room here. For Part (b). Now Part (b) is asking for the final velocity the magnitude and the direction of the final velocity. So remember if you want to draw velocity it is sort of tangent to the paths so it look something like this. And it is a 2-dimensional vector. So it is composed of Vx and Vy and the way you're going to get your velocity is by finding Vx. Your final Vx and your final Vy. I need those tools to be able to find Vfinal and to be able to find theta final as well. OK so let's do that. I need to if I am looking for V and theta finals I need to find Vx final and Vy final as always Vx straightforward Vx never changes Vx is just 4. But I do have to find the Vy final which actually because I avoided the quadratic equation and I went through the tool equation process here. I already know what the final velocity is, the final velocity is negative 8.23 meters per second. So I already have what I need. I have the two components of the vector. I just have to put them together. The magnitude of V is the Pythagorean of its components. So Vx square plus Vy Square. So, the numbers to plug in here are Vx 4and Vy is this, ok. And if you plug them carefully here, you get a V, I got a here of 9.15 meters per second. That's the magnitude of the velocity and the angle is the arctangent of y over x. Here I'm going actually plug in the numbers the Arctangent of Y which is negative 8.23 divided by X which is four. And the answer you get is 64.1 degrees negative rather 64.1 degrees. That's what your calculator will give you. And that is the final answer because remember with the arctangent you have to be careful this has to be either in the first or fourth Quadrant for you not to have to touch up the arctangent.This is in the fourth quadrant so this angle is good and it requires no further work. OK, so now there's a similar one here and I want you guys to try this practice problem and hopefully you get it.

Practice: You throw an object from the top of a building with 50 m/s directed at 53 o below the horizontal. It covers a horizontal distance of 80 m while in the air. Find its: 

(a) total time of flight.
(b) final velocity (magnitude and direction).