5. Projectile Motion

Positive (Upward) Launch

# Using Single Intervals in Positive Launch Problems

Patrick Ford

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Hey, guys. So by now we've gone through all the steps to solve project on motion problems for any given problem, we're just going to sketch the trajectory and draw its past in the X and Y. And then once we figure out our target variable, we're gonna pick our intervals and then start using our equations. And what we've seen in Project on Motion is that oftentimes you can choose different intervals and still get to the right answer. For example, imagine if I had this upward launch here to some lower height at Point D, and I wanted to talk to calculate the total amount of time from initial to final. We'll actually have a couple of different paths that I can choose from a couple of different ways I can solve for that. First, I can try to go from a to B calculate that time. So Delta t from A to B and then added to the time that it takes to go from B to D. So just adding those two smaller times there. But another option is like I also just try to calculate the time from a to see because I know that is symmetrical, and I can try to use some of the symmetry points there. So that's Delta A from Delta T from A to C and then attitude Delta T from C down to D. So if I had enough information, I could actually solve and get the same time and same answer using both of these different approaches. But oftentimes, the easiest approach to solving these problems is just by using a single interval, going from a all the way down to D, you could absolutely do that. And in fact, that's gonna be your best bet on solving these kinds of problems. Usually you should try as best as possible to solve these problems using a single interval, like from A to D, because it's gonna make your life a whole lot easier. Let me show you how this works in this example down here. So we've got a cannon and we've got the initial speed in the angle. We're gonna calculate the vertical component of the velocity at the point where it hits the ground. So let's just go ahead and go through the steps. First step is we're gonna draw the past in the X and y. So we're gonna go ahead and have passed in the X and y. That's point B, C and D and the Y axis. It goes up and then back down again. So it's B, C and D. And so now that's the first step. The second step is we're gonna figure out the target variable. So in this, uh, first part here, we're gonna calculate what the vertical components of the velocity at the point where it hits the ground. That's point D. So remember, at any point there's two components of the velocity. There's VD X. There's the X component and the white components. That's V. D. Y. And both of these combined to form a two dimensional vector using the Pythagorean theorem V. D. So which one of these variables we're looking for? We're looking for the vertical components that is V. D. Y. So that's our target variable. So now just comes to what equation or what interval are we gonna use? Well, remember that these problems were gonna try as best as we can to use the interval from a D. D. If we have enough information, if we get stuck and we turns out we don't. We can always just come back and then picked different intervals. So let's try to use the interval from A to D. All right, so I'm looking for a y axis variable. So that means I'm just gonna use I'm gonna list out all of my wife variables. Right. So this is gonna be a y, which is negative. 9.8. And that's regardless of the interval that we're using. Remember a Y. It was always negative Jeans negative 9.8. No matter which interval you're gonna choose. So our initial velocity is our velocity at point A. Our final velocity is V D y. That's actually we're looking for, You know, we've got Delta y from A to D and then we've got t from a to D. So let's start knocking down these variables. So I'm trying to figure out v a y. And that is I configure that as long as I know the initial velocity, which is V A, which is 100 the angle which is 30 degrees, I have both of those magnitude and angle so I can calculate the components my V not X, which is V a X, which is just Vieques throughout the entire thing the entire motion will never change is times the co sign of 30 and that's 86.6. That number is never gonna change throughout the whole motion with y Velocity, on the other hand, will change. That's V A y, and that will be 100 times the sign of 30. And that's gonna be 50. So that's our variable. There we have V A Y is equal to 50. So which other ones which other variables can we solve for? We've got Delta y from A to D and then t from a d. D. Well, the only other thing that we're told in the problem is that this cannon is fired from a m cliff so that what that means is that this height over here from the cliff down to the ground is 40. And if you think about it, this is actually just the vertical displacement from a down too deep. Forget about what happens in the middle. So your interval is going from a all the way down to D. But all you really care about is the initial and final And so because your displacement in the vertical is just gonna be 40 downwards again, forget about what happens in between. It doesn't matter. That goes up and then comes back down again. It's gonna be negative. 40. So that's negative. 40 over here. And so therefore, we have our three out of five variables. I've got 12 and three, and I'm gonna pick the equation that ignores time. So that's equation number two. So I've got V d Y squared equals V A Y squared plus to a Y times delta y from A to B I have all those numbers and I'm just go ahead and plug so my V d is gonna be the square roots of. And then I have 50 squared plus two times negative, 9.8 times negative, 40. If you go ahead and work this out, what you're gonna get is you're gonna get plus 57.3 and also minus 57.3. Because remember that whenever you take the square root of a number like the square root of nine, there's two answers positive and negative. Three. Both of those things, when you square them will both equal nine. So which one of these makes sense? Well, a positive 57.3. Meaning it would mean that the velocity is going up and a negative 57.3 meetings means that it would go down. So obviously at point d r velocity component is gonna be downwards. So therefore, this is not the right answer. And this negative 57.3 degrees is our right answer. All right, let's let's move on. So this point, part B Now, now we'll be looking for we're looking for the total time t from A to D. So we go through the steps again. We've already have our past in the X and y. We've already determined the target variable and the interval were still just gonna try to use a two D. In fact, T a d was the was the variable that we initially ignored in this first part. But now that we know V. D. Y, now we know this is just negative 57.3. Now we actually have four out of the five variables and so we can actually use any one of the equations to solve 40. Any one of these equations will be, uh is totally fine for solving the time. So what we're gonna do is we're gonna use equation number one because it has the least amount of terms. It's the simplest equation to use. So this just says that v. D. Y here in our same interval at a D is just gonna be v a y Plus on this is gonna be negative. Whoops. Plus a Y times t eight a d. So we've got negative. 57.3 equals 50 plus negative 9.8 times t a to d we go ahead and move this over, we're gonna get negative. One of 7.3 equals negative, 9.8 times ta two d. And so we're gonna get negative 107.3, divided by negative 9.8. The negatives will cancel, and then you'll just get tea from A to D, which is equal to 10.9 seconds. So notice how we were able to solve all of these problems here both of these parts just by using a single interval. It's gonna make your whole life a lot easier if you can try to do this is best a zoo much as you can. Anyway. That's it for this one. Guys, let me know if you have any questions.

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