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Intro to 2D Motion | 32 mins | 0 completed | Learn |

Projectile Motion | 67 mins | 0 completed | Learn |

More Projectile Motion | 42 mins | 0 completed | Learn |

Initial Velocity in Projectile Motion | 20 mins | 0 completed | Learn |

Circular Motion | 27 mins | 0 completed | Learn |

Additional Practice |
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Motion in 2D |

Motion in 2D & 3D With Calc |

Projectile Motion: Horizontal & Negative Launch |

Projectile Motion: Positive Launch |

Projectile Motion: Launch From Moving Vehicle |

Relative Motion |

Concept #1: Launch Up From Height

**Transcript**

Hey guys we're now going to look into a type of projectile motion problem where we launched something up at an angle and then it lands below your original height. So something like this. Your final height is below your original height. All right. So I call this launch up from a height not very creative. So in these problems the first part of the motion is symmetric. And what I mean by first part is from A to C, remember if two heights are the same if two points are the same height, they're symmetric that means the time to go up is the same as the time to go down. So by first part. I mean A to C. But the object drops further drops further is C to D. This piece over here and that's what's different between this type of problem and let's say symmetric launch problem. OK. So when we're solving these problems there's there's three legs here AB, BC and CD. So we're gonna get to pick intervals for some of the things we're solving for and whenever possible we're going to try to include Point B in our intervals. OK so if you have a choice of intervals sometimes you do sometimes you don't. We're going to try to include Point B and I'll show with this in this example. Point B is simply the highest point the max height. And the reason we want to include this is because the velocity at point B in the y axis is zero which means our equations will simplify there'll be a little bit simpler to have one last term and that's good news. So show you how that works. Lets go and do this problem. And here it says you throw an object from the top of a 100 meter tall building. So this is 100 meters with 50 meters per second so Va equals 50 and an angle of 37 degrees above the x axis so that's this I'm going to call this theta a equals 37. Now the first thing I'm gonna do here is actually decompose Va into Vax and Vay. So Vax is which by the way Vax looks like this is simply going to be my Vx because the velocity in the x axis remember stays the same always. So Vax the same as Vx, it is Va cosine of theta a and you gets 50 cosine of 37 which is 40. That's positive because it's going to the right so that makes sense that you get a positive. For here Vay is Va sine of theta so 50 sine of 37 which is a 30 meters per second. So Vay is 30 and it's a positive 30 because it's going up we don't have a lot of space. So I want to try to compress all the information here. Of course that's the first thing we do just put that up real quick. Now let's hear what they're asking us asking for the maximum height. As you look three or five variables in motion you should identify that the maximum height has to do with your Delta y. But it's not just her Delta y let's say from A to B but instead it is this original height here one hundred meters plus the height from A to B. That's your total maximum height. OK. So your maximum height will be 100 plus the this length here which is your Delta y. From A to B. I have the 100 obviously but I don't have the Delta y, we will have the final answer for the maximum height. So that's what we have to do. Let's go find Delta why between A and B. Now remember for every question you're asked for every variable you're looking for. You have to go through the steps, determine the axes right. So Delta wise obviously the Y-axis. Pick an interval. Well if I'm looking for Delta y from A to B the intervals are picked for me. It's gonna be A to B and pick an equation that part you do have to do. So to make this a little bit simpler rather than work with the entire picture I mean sort of cut out just this part just the Y-axis from A to B would look like. And again on the y axis you're either going up or down. In this case from A to B I'm going up. A to B. The initial velocity here is Vay which is 30 and the final velocity here Vby is zero. Because that's what happened at the maximum height right, Vby over here is zero. What else do I have I'm looking for Delta y for this interval. The other two variables are delta T, I don't have Delta T and acceleration. Acceleration is gravity but it's negative because it's going down. And remember that going up is positive and going to the right is positive. That's all we're going to use as a standard. Delta y is going up so it should get a positive. This velocity is going up. So this is a positive. And I don't know time but that's OK because I already know three out of five variables. So time is simply my ignored variable. But I was still able to find delta y since time is my ignored variable that means I'm gonna use the second equation to solve this.

So the final velocity squared equals the initial velocity squared plus 2a Delta x. But when the y axis and the final velocity the y axis is zero. The initial velocity on the y axis is 30. The acceleration is negative 9.8 and Delta y is what I'm looking for. OK. And if you move things out of the way you get that move things around to solve for Delta y you get the Delta y is forty five point nine, I'm gonna round this to forty six meters. So Delta y is 46 meters but remember Delta y is not what I was looking for. I was looking for Delta y so I can plug it in here. And what I was actually looking for is the final, the maximum height which is 146 meters. So that is actually my final answer right here. Maximum height is 146. Delta y is just to step in between now the next part that's part (a) next part is asking for the time to reach the maximum height. So the time to reach the maximum height so that's. That's Delta T. And again to reach the maximum height that's A to B. So the intervals already picked out for me A to B. Now what's special about time is that time could belong to either the X axis or the Y axis. And typically I would start with the x axis but I'm going to just advance you let you know that if you try the x axis it's not gonna it's not gonna work. So I'm gonna just jump into the y axis and actually if I'm looking for time from A to B all it is this guy right here my what used to be my ignored variable. Now I can solve for it. All right so except that now I actually already know four things about this motion which means I'll be able to pick not just one equation but two equations. So we'll pick the simplest one. Right. In this case I can use the first equation and Vfinal equals Vinitial Plus aT. The final velocity we're talking about the y axis as we go up, the final velocity is 0. The initial velocity is 30. The acceleration is negative 9.8 T. And if you move things around you get that T is three point zero six seconds. Alright. In this is the answer to part B. That time is three points zero six seconds. All right. So part C asks for the total range and we call that another word for that another term for that is horizontal displacement. That's just a Delta x. So let me do this over here. I'm looking for Delta x. Obviously, delta x is an x axis is variable and a good thing but x axis again there's only one equation. So letÕs write that. And it says delta x equals VxT. Vx is always the same in this case it's 40. 40T. OK. Now I don't have T. Because if you're careful here this delta x is actually the entire right the total range the entire range from A all the way to D. So if I want delta x from A to D Vx is always the same but I need to use the time from A to D. And I don't have the entire time from A to D. I just have the time going up. So that's what we're going to have to find first. OK so I'm gonna have to first find what is delta-T from A to D and then I'll be able to plug it back in here. In doing this. I am already determined that intervals gonna be from A to D and since I was I just got stuck in the x axis we're going to do this in the y axis. OK. So I'm going to move over here and find what is the total time in the y axis. Actually we do a little bit more work here. The time in the y the time in the y axis from A to D again the time is the x axis or y axis y doesn't matter. But we're solving this in the y axis. You could think of this is the time to go from A to B plus the time to go from B to C plus the time to go from C to D. And if you look closely you have this first time 3.0 6. But if you know that it takes 3.06 to go up. You know that it also takes 3.06 to go back down from B to C. Right. You know that this time here is also three point zero six Right. So you actually know both of these. One way you could do this is just finding the time between C and D. Right just write a little interval diagram from here to here and find find the time in the y axis. The thing is it turns out that if you go from B to D it's actually easier and it's the point that I made in the beginning that as much as possible we want to try to include. We're going to try to include point B because things were simplified. OK so we're gonna instead of thinking of it as AB, BC and CD.

I actually want you to think of time total any kind of time total as simply the entire time up plus the entire time down. In other words this is AB and this is BD. And if you do it this way. Notice that this guy has a B in it and this guy has a B in it. And that is better. So I want to use the entire time to fall. OK. I already know this. So this is actually what I need. So let me go over here to the side and find what is the time delta T from B to D. And again we're going to do this in the y axis because we're stuck in the x over here. Pending what we find in the y axis. OK. So that means I gonna draw a little diagram from B passes C ` D down here and we're going to plug in all the information we know with this in the y axis the initial velocity is Vby and that's zero because it's the highest point. The whole reason why we're doing this from B to D is because we want this to be one of these numbers to be zero. This is Vdy which I don't know. The time is what I'm looking for. The acceleration is gravity. Now it's negative gravity negative 9.8 because it's going down. What else the fifth variable here is Delta y we do have this but you have to be careful if I'm going from B all the way to D. I have to use the entire height which is not 46. Remember this is 46 over here. You don't use the 46 you donÕt use the 100 but you use 146. The entire thing and you falling so it's negative 146 because you're going down and going down is negative. So if you look around you know three out of five things. And this is your ignored variable. So I'm going to use the third equation to figure this out. Right. So let's use a third equation here. The delta x equals VinitialT plus half of aT square. Delta x is actually Delta y which is negative 146. This is zero because we're talking about the velocity in the shoe on the y axis that was zero. This is half negative 9.8 t squared. OK. And then if I move some things around and I solve for T I get T equals the square root of some stuff. And if you plug that into the calculator you get five point forty six seconds ok. Five point forty six seconds. That's not the final answer. That's just all we have for now. Let me make a quick point here. Notice how you're looking for time. And this guy canceled hear, it canceled because we're going from B to D and the velocity of B, this is Vby is 0. If you want going from C to D this guy would not have cancelled and I would have to use the quadratic equation to solve this which sucks right. So we want to avoid that as much as possible. So you want to include the point B. So I got this time and I needed this because this time is actually this guy over here. So time total is three point zero six plus five point forty six. And I have this here the total time is eight point fifty two seconds. Now all of this we weren't actually directly looking for the total time. All of this is so that we can plug this total time in here. OK. Delta x total is Vx which is 40 times the entire time. Eight point fifty two. And if you do this you get this round to 341 meters. Ok so this question is long and tricky because it asks for your total Delta x he never asked you for the totality but you had to go get it. OK. So that is part C. Maybe a little space here. And now we're going to go to part D. Part D asks for the magnitude and direction of the velocities at point B C and D. OK. So first Vb remember the vector the velocity vector is tangent to the path. So it looks kind of like this. So at Vb which is your highest point Vby is always going to be 0 and Vbx remember the velocity in the x axis never changes so Vbx just Vx. It started 40 continues at 40. If I want Vb I need to have Vbx and Vby. Every vector for you to find the magnitude of a vector you have to know its components. And then I can just do the Pythagorean theorem with the components, here notice what happens my Vby is zero. So Vb is really just Vbx which is 40 ok. That happens because this vector is entirely in the horizontal axis. It has no y component right it's flattened the x axis. So the entire vector is the velocity in the x axis. And because of that theta is simply 0.

Now if you want to verify that you could have done that theta at B is the arctangent of y over x so 0 over 40 and that would have been zero. But just by looking at the diagram you can see that and the velocity. The angle for the highest point of the velocity at the highest point will always be zero because it's always going to look like that. OK. Now let's look at the C again to find the C I'm going to need the Vcx and Vcy. Vcx never changes it's a 40. Now Vcy, why is a point that is symmetric to A. So if my velocity going up Vay is 30. My Vcy is negative 30. Ok so Vcy is negative 30 because of symmetry. And I can now having these two numbers I can now get my Vc which is going to be these two. So forty square plus negative 30. Squared and you get a 50 meters per second if you do the arctangent theta C, arctangent of Vcy. 30 over 40 you get a negative 37 degrees. These numbers again shouldn't be a surprise. These numbers are the same as these numbers here. OK except that the angle is different. The magnitude of Vc is always going to be the same as the magnitude of Va. If A and C are symmetric and the angle of C is always going to be the negative of the angle of A if the those two points are symmetric and that's because the sides the legs of the vector are the same right they just flipped. This is my X this is my Y. And then at points c it does this and just a vector over here. Just flips down instead of up. But the magnitudes are the same. So. We actually didn't have to calculate anything. Vd is going to be a little bit more complicated. Vdx is 40. But I do have to find the Vdy to be able to do this. Ok so we don't have Vdy we gonna have to go get Vdy. And again if you want Vdy which is this velocity down here. Right. Vdy, you have to write the equation you know pick up pick an interval write an equation. And the best interval here is going to be again not C to D. I know that this is 30 so maybe I could just go from 30 and then figure out what D is but it's actually better to use B all the way down to D. Once again we weren't involved B as much as possible. That's one reason to do it because B the velocity at B is zero. But also because we only drew this over here and we have a bunch of numbers already kind of worked out. In fact we already know four things about this interval. I already know that this time is five point forty six so I'll have a greater selection of equations if you will. So let's go and look for. Vdy which is this guy over here. OK. So I'm going to go over here and find Vdy. I can simply use the first equation. And Vfinal equals Vinitial velocity on the y axis and the initial velocity they are zero. And I can just plug in these numbers negative 9.28, the time I have to use is the time for this interval. So it's five Point forty six. Not any of the other times. And if you do this. You get. It here somewhere. Fifty three point five negative fifty three point five meters per second. That should make sense that it's negative because this thing is going down. So we expect that it will be negative. OK. So now I know that this is negative. Fifty three point five right here. And I can just do Vd Pythagorean. If you plug this you get sixty six point eight. And if you do the angle the arctangent of Y negative fifty three point five over x 40 you get fifty three point two degrees, a negative fifty three point two degrees and that is the correct angle because Vdy over here once again you're going this way you are this is your Vd. You are in the first fourth quadrant. So you don't have to touch up angle. The angle is the correct one coming from the arctangent. All right. So these are all the numbers Vb is 40, Theta B is zero. Vc for some reason I change colors here and then Vd.

So I know this is a long example but this sort of helps me explain all the little details about these kinds of problems. There's one more thing that I want to show you. So it says here if we're answering multiple questions like which is did you go from A to B and B to D you do all kinds of things. It's often the easiest or best to break up the motions so in this case I did AB and then we did BD. And once again notice how that ends up B being part of both. That's good. Usually that's what you want to do. You have to break it up because you have to find parts of it anyway. But if I only ask you for one thing. Or in some other situations sometimes it is better or easier or faster or whatever the situation to actually just solve this in a single step going from A to D. And that's what I want to show you real quick. So you know that you can do this here we're going to throw a rock from a 60 meter cliff. So you make a tiny little cliff here, 60 meter. We're gonna throw this rock with 50. So I call this Va equals 50 at an angle of theta a equals 53 above the x axis. And obviously this thing is going to go and land over here somewhere. I'm going to call this A, B, C which in this case is not important. All right. And I want to find the vertical component of the velocity at point D. So the Vdy. What is Vdy And I want to do this in a single step. OK. So let's do that real quick. So. I'm looking for a velocity in the y axis so I'm going to use a y equation and y diagram and I want to go from A all the way to D in one shot. So let's draw what's happening in the y axis here from A to D. You're going from A you go up to B and then back down to D, ok and there's B here and there C here. Now if I'm gonna do this in one shot I mean I have to pick one equation and do it but it's going to be easier if I actually list my variables real quick. I have my initial velocity which in this case my velocity at Ay which I have to find by decomposing. And then I have my. Yeah that's my initial velocity right there. And then I have my final velocity which is my velocity at point D, Vdy which is what we're looking for. OK. Then there is the acceleration which is gravity negative nine point eight. Because we're going down. Then there is the amount of time that it takes delta-T. And then there is the height. So I can find Vay. Let me do that real quick Vax equals 50 cosine of 53. Which is 30. This is just my Vx throughout. And my Vi or Vay is 50 sine of 53 which is 40. So this initial velocity here is 40 up so its 40 positive. Okay. Now I'm looking for this variable which means I need to know two others or three others. I have one two the third variables have to be either time or height. If you look a question B its asking for the total time of flight so I probably not supposed to have that just yet. Do I have Delta y. I do. Now this is where it's a little tricky because I am on top of a 60 mirror Hill. And I go up and then back down. But the way to approach this is to look at the very beginning and then the very end, forget what happened in the middle. And if you look at the very beginning I start here the very end I go here. So effectively what's happening is I'm going this way beginning to end forget what happens in between and going down. So I can say that my next Delta y is that I'm actually beginning to end falling 60 meters. Right. Again forget the middle. I'm falling 60 meters from the very beginning to the very end. So it's negative because it's going down. That is my third variable and I can now use delta t as my ignored variable and I can use my second equation to get our final velocity which is this variable here. OK. So the final velocity is the initial velocity forty square plus two Negative 9.8 and then instead of delta x It's actually Delta Y which is negative 60 ok and if you plug in all these numbers and you carefully solve for V you get that V is a square root of some stuff. This is your final velocity which is I have it here. You get a fifty two point seven meters per second. But be careful when you're coming back down your velocity Vdy is going down so it should be negative. So I'm going to put a negative in front of here. Remember when you take the square of a number. For example square of nine you really get a plus or minus three because three square equals nine and negative three squared equals nine which means you can just add this negative here and you're not violating any crazy rule. Right. So that is part (a) that the final velocity in the y axis Vdy is negative fifty two point seven and I did that in one shot. I didn't go up and then back down to that in one shot. All right. Now let's do Part B let's try to squeeze part B over here. What is the total time of flight. So again in one shot. So I can. It's the same interval from A to B in here I can just use the first equation Vfinal equals Vinitial plus aT, Vfinal again in the y axis we're going up and down so its Vdy, Vinitial is Vay and the acceleration is negative gravity and time is what I'm looking for. The final velocity in the y axis I have it right here we just got it negative five two point seven. The initial velocity in the y axis was 40 minus nine point eighty. And if I shuffle everything around I get that T equals four or nine sorry nine point four six roughly nine point four six seconds is my total time. OK. So we just had two questions in two very different ways. One of them asked for a lot of stuff so we broke it up and that's usually what you do most of the time because these questions are usually an ask for a lot of things. But in some cases you might want to just do it in one step and get it out of the way quicker. All right. So that's it for this one.

Concept #2: Launch Up To Height

**Transcript**

Hey guys we're now going to talk about projectile motion problems where you throw something up at an angle but then it hits its target before coming back to the ground so let's check it out I call these launch into a heights because your final position is higher than where you are still launching into a specific height in these problems we're going to usually use a combination of X. and Y. equations like usual but there's going to be a focus on this third equation here the third equation of motion for the Y.axis and that's because the variable Delta Y. will always be of interest in these problems right.So this is Delta y this is my V initial in the Y. axis and this is my A in the Y. axis which obviously in the Y. axis is G. negative because we're going down so I think this is best explained some problem some examples so let's check this out a fireman sixty meter away from the building so that's your delta X.Shoot water from a hose at thirty meters per second so the water comes out of here at thirty meters per second and at an angle of fifty three. Degrees above the X. axis so that's the information I know my V.A. and my theta A so I can first of all decompose this for a quick So it's true that the X. which is the same as my V.X. everywhere is V A cosin of theta A so it's thirty cosine of fifty three degrees and I have this here it's eighteen meters per second.

So this decomposes into eighteen in my V.A. Y is thirty sine of fifty three which becomes a twenty four so this velocity going up here is my twenty four cool so the first question is what is the time that it takes to hit the building so if I call this. A this point B. And this point C I'm looking for the time from A to C. Notice that from A.to C. in this situation is not symmetric.So I cannot use any kind of simplification that I could have used in symmetric problems OK.So to solve this I'm going to do what I usually do which they start in the X. axis and then I go on the Y. axis and you know I want to give you guys a systematic way of solving these so just start in the X.axis See what you get if you get stuck you go to the Y. axis right so X. axis delta X. equals V.X. T. This is the entire time from A to C. I know delta X. um So if I want T. I just have to know both of these and actually know both of them right so time is delta X. over V.X. so delta x is sixty V X. is eighteen.AND If I divide these two I get that the total time is three point three three seconds so is actually able to get time from the X. axis remember time could be coming from the X or from the y I try the X. first because it's simpler and hopefully it works and it did OK so for Part B. I want to know at the height and which it hits the building so I want to know Delta Y..From A to B..So I was going to go to the Y. axis the Y. axis looks like this if I were to draw it this thing goes up to B. and then it hits C over here and B. in C. I'm going to use this third equation because I have all the information already right I'm looking for Delta why. I know the initial velocity the I know time I know acceleration I know. Time so I can easily usethis equation Delta Y. equals the initial velocity in the Y. axis is the initial velocity in the Y. axis twenty four. The time is three point thirty three.

Plus half my acceleration is actually negative nine point eight because my acceleration is negative G. and G. is positive nine point eight delta Y. becomes negative one point eight.Time square ok. So I actually do know the times of three three three point three three squared you plug in all these numbers and you get your final answer you get that this is eighty minus fifty four point four which is twenty five point six.In that is it that's our time and that's our Delta Y OK So this is a straight forward version of this question of this type of question I want to show you one more case thatwe have with these right so in some other cases we're actually not going to know if the object is going to hit its target on its way up or on its way down in the previous example I jus the diagrams you knew was coming the way down but if you shoot something its path looks like this and if you want to know for example how long it takesto get ten meters wide you get a ten meters here. Or I could get a ten meters here so you could take T one or it could take T. two there are two possible types another situation is what we get here in this example number two where if you want to hit a building at a certain height you could do it from this distance or from this distance so.It's because and that's because here I'm hitting on the way up so it's faster right so it takes t one and I can do this from a distance delta X. one. But here it's going all the way up past that point Max height end it's way back down so it's going to take time T. two or the distance delta Xtwo. that's what this example going to deal with OKso it says here that the water's coming with initial velocity of twenty five and an initial angle of thirty seven.

and you reach a building ten meters above the ground so your Delta Y. from A to B. equals ten all right I'm going to call this a and B. because usually I would do A B. and then that C. but I could just be going from A to that point this point could not even exist I don't know so i am just going to skip that and just call this B OK We're just going to have one simple interval beginning to end and irrespective of whether I'm going up and then back down in the building or just hitting the building on the way up my Delta Y. in both cases just ten OK. So it says how far from the building should you position the hose so it's asking for Delta X. one and delta X. two it says you there are two possible distances in the problem going to kind of walk us through that first we are going to find the time and then we are going to find the two distances since I'm looking for two times this is a big hint that whenever you're looking for two T's it's a big hint that you should be using equation number three and that's because equation number three sometimes will give you a quadratic will end up in a quadratic and you remember the quadratic equation will give you two values of T. so we should be expecting to get the quadratic right so let's get started we're looking for the two times that this would take some just going to use this equation now that I know that this equation is the way to go I don't have to go to the X. axis tried that and come back to the Y. axis some just going to go straight to the Y. axis and use this equation but instead of delta Y this is the initial velocity of the Y. axis and this is the acceleration on the Y. axis which is negative G T. square now just a matter of finding time from here I should have all the numbers delta y is ten going up V.A. Y I can find that I can decompose V.A. vx equals V.A. twenty five cosine of delta thirty seven so this is twenty and V.A. Y. is twenty five sign of thirty seven and this is fifty K. so V.A. Y. is fifteen. Time that's what we're looking for this is minus four point nine T. square now remember the quadratic equation it should take the form of eighty A T square plus B. T plus C. equals zero and A is a number that goes in front of the T. squareand c is the number that has no T attached to it. the one big rule here obviously this has it was your but the one big rule is that A must be positive Other than that it doesn't really matter it must be positive so I need to move this over to the left so make a positive in now I'm going to move this over to the left as well so that the right side is empty and I get this.

So now my A B. and C. A four point nine B. negative fifteen C. equals ten we're going to plug into the quadratic equation.Which is T. equals minus B. plus or minus the square of the square minus 4A C. divided by twoA let's do this slowly minus negative fifteen become a positive fifteenplus or minus the square root of negative fifteen squared. Minus four. A.C..All of that divided by two a two times four point nine And if you do all of this and you simplify you should get at the first number here first term is fifteen plus or minus when you work out this entire thing here you take that square root of that you get a round of the five point four divide that by nine point eight once you get here remember you're supposed to get two times the plus or minus is going to give you two times one the first time you're going to do using the negative sign and then the second time when to do using the we're going to find use a positive sign it doesn't really matter which cult you want to teach you right so T 1 is going to be fifteen minus five point four divided by nine point eight and that gives us Point ninety eight seconds and T 2 is fifteen plus five point four divided by nine point eight and that gives us two point zero eight seconds those are the first two times the first times and this is the answer to our part A. That the two times that it takes part A Point ninetyeight. And two point zero eight for Part B. is just a two distances so that you don't access.The outside X. is V.X. T. So if I want delta X. one it's just V.X. which is always the same T one so V.X. is twenty times nine point eight and for delta X 2 this is going to beV.X. T two twenty times I'm sorry this is not nine point eight point nine eightAnd then this is true point zero eight.All right.and the answers are nineteen point six meters as well as forty one point six meters obviously. This one is if you hit it on the way up.Short distance and this is if you hit it on the way.Down OK but remember the big thing here is I'm going to use that third equation to figure out how to work this right that's it for this one.

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