Initial Velocity in Projectile Motion

Concept: Finding Initial Velocity in Projectile Motion

11m
Video Transcript

Hey guys we're now gonna talk about symmetric motion problems where you don't have the initial velocity Let's check it out. So usually in symmetric motion problems or projectile motion problems in general you're given the initial velocity the initial angle and then you're asked to find the range how long it's in the air for and all that kind of stuff. But what happens if you don't have the. So in specifically for symmetric launch remember symmetric is where you start and you come back to the same height. When you're missing either Va or theta a or both. We're gonna solve this using a combination of two to three equations. Now the way we're gonna solve this a little bit different from the problems where I am giving you Va or theta a, so we are gonna use a different strategy. Remember our vector equations right, so two of them have to do with decomposition. How to get the Vax, Vay. But two of them have to do with how to put a vector back together. The magnitude of any vector is the Pythagorean of its components. And the angle of any vector is the arctangent of its components like this. So the idea is if I'm looking for Va and theta a I'm actually going to find Vax and Vay and then use the components to find the vector and the angle. So that's going to be our strategy find Vax and Vay, remember Vax is simply Vx because the Vx never changes. OK. Now to do this I'm going to derive three new equations for you. Not new but simplified versions of our motion equations that I think are going to be very helpful. OK. So if you write Y equations in the BC interval. So this is A this is B and this is C so BC is this. If you write equations in the BC interval you're going to get very simple versions of those equations because the initial velocity is zero. And what you're going to have the simple combinations of all of your three equations and I'll show that in a second. So you and I have that in such a way that if you know any of the three y variables you're going to be able to find the other two using these three very simple equations. OK.

Again if I go from B to C the initial velocity here Vby is zero. And that's why we're going to use BC and not AB or AC or whatever. And if you notice your 3 motion equations they all have the initial Right and this is my Vinitial Because I'm going from B to C so Vfinal equals Vinitial plus aT becomes just V equals aT. Vfinal squared equals Vinitial. Square plus 2a delta x becomes just this and delta x equals Vinitial T that cancels plus half of aT square. So these equations simplify a lot. Now to avoid problems with signs and directions and all that kind of stuff. I'm going to try to simplify this even further so if you're going from B to C just for the sake of deriving these equations I'm going to say that going down is positive so all these numbers will be positive. OK so what I have is and I'm also going to call and set up Delta Y just one more thing instead sort of Delta y I'm going to call this H and just say you know what let's just say H is the height whether it's positive or negative we're going to use that as a positive number because the height is positive. So we don't have to worry about signs what that's going to do is that all these numbers all these variables all these values and these equations will be positive. So let's work this out. This is the velocity going down which is gravity going down as positive grab is going down so this is positive. And this is the time but it's only the time to go down, okay. So the second equation is the velocity going down squared equals 2 gravity again as a positive. And instead of Delta x is Delta y, Delta y going down is positive in this case but I'm just going to call this H. Okay. And three is Delta y or delta x becomes out to Y but I'm going to call it H, equals half of gT squared. And this is the time to go down. All right. So here's my point that the these equations will be simpler but there's a few other things here that for example T up equals T down. So I'm really solving for either one of the two times. But it is not the total time the total time is 2T up or 2T down. And another thing is that the velocity going up has the same magnitude as the velocity going down. So these equations could very well have been written as Vup or Vdown Tup or Tdown. So I'm actually going to write the sort of final version of these equations. I want to write just Vi equals g.T. Now I'm going to put a T up here. Just so we're sure that's so we know that that's not the total time. The why whether it's up and down doesn't matter. A Or C equals 2g.h. And the last one is going to be H equals half of g.T up square, again doesn't matter that I'm using time up or time down. They are the same. OK and remember that the time the total time is a combination is the addition of both such as twice the time up or twice the time down. So this is a simplified set of three equations that I'm going to be able to use. And with these with these as long as I have any Y variable I'm going to be able to find the other two if you look here I have V and T this has V and H and this has H and T, so I have every possible combination. Let's do an example that I want you guys to do a problem. OK. And remember this strategy is going to be to find Vax and Vay and then use those to find Va and theta a. So here an object is launched from from the ground at an angle like this. It reaches its maximum height and then returns to the ground. Because it returns to the ground. This is a symmetric launch which is a type of problem we're looking at. I'm going to call this A, B and C. And this is my Va, this is my theta a which is what we're looking for. But again the strategy is we're actually going to look for Vax and Vay first.

So I'm going to go for Vax first because the x axis is simpler. So there's only one equation delta x equals VxT. So Vx is simply Delta x over T. I have Delta x right this problem says that you go a horizontal distance of 240. And by the way it also says that the height is 45 something we'll call this H equals forty five. OK. So I have 240 divided by time. The time I don't have. But I do want to point out since since I'm using 240 which is the entire width of the motion it's the entire horizontal displacement of the motion. I have to use T total, in these equations here which we're going to use and second are only going to give me half of that time. So remember that T` total is twice the time to go up. All right so now I'm stuck I'm stuck on the y axis I can't do anything else. I am sorry I'm stuck in the x axis I can't do anything else so I'm going to go to the y axis looking for time, ok and remember this is one of the big points here. If I know any of these three variables I can find the other two. And here I know my H. So from H I'm going to be able to find the time to go up one of the time to go down it's the same thing. And from that I'm going to be able to find the time total. OK. Which is what I want for this so I can use the third equation here very straightforward, H equals half gT square. Again these equations all these numbers are positive because of the way that I simplified them in their derivation process. And I can just find T, T is the square root of 2 H over g. And if you plug all these numbers in here you get three point zero three seconds. But that's only my time to go up or the time to go down. The total time is six point zero six seconds. OK. So to find my velocity let's move this over here to find my velocity its 240 divided by 6.06. And if you plug in these numbers carefully you get thirty nine point six meters per second. OK. So I found Vx. Now I just have to find Vi. Vi obviously sits in the y axis. So I'm gonna use one of these three simplified equations ands to find V, I can use either the first or the second. I'm gonna use the first one here so Vy equals g time up. So g is 9.8 the time up is three point zero three in Vi therefore is twenty nine point seven. OK. Now that I have these two velocities I just have to piece it together and find a vector. OK. So Va is the square root of Vax Vay. Or the Pythagorean between the two and these velocities are thirty nine point six squared the axes and change Vi here is my velocity going up which is the same as the velocity going down. So I can use that. And you plug this all in and you have forty nine point five meters. And then the last thing I have to do is find the angle theta a. It's just the arctangent of y over x. So the arctangent of y over x and I gets the rounds up to thirty seven degrees ok. So just to wrap up real quick strategy find Vx, find Vy when we're looking for Va and theta a and I can use this simplified set of equations that allows me to very easily navigate between the three y variables that I'm gonna have. All right. So that's it for this one.

Problem: The range of an object launched from the ground at an angle is 300 m. If the vertical component of its velocity just before returning to the ground has magnitude 50 m/s, find the magnitude and direction of its initial velocity.

4m

Problem: An object is launched from the ground with an initial 40 m/s at an unknown angle. It reaches its maximum height in 3.5 seconds, before returning to the ground. Find its initial launch angle.

3m

Problem: An object is launched from the ground with an unknown initial speed directed at 30° above the horizontal. If it reaches a maximum height of 40 m before returning to the ground, find its initial speed.

3m