Hey, guys. Let's check out this problem. We're playing catch on a faraway planet. We've got some information about the launch speed and angle of a projectile and the distance that it travels and we're going to calculate the gravitational acceleration on this planet. So let's just stick to the steps. Basically, what I've got here is I've got level ground. I'm told that I'm throwing a ball upwards with some velocity at 37 degrees, and then it's going to later return to the ground again. So it means when I draw out this trajectory, it's actually just going to be a symmetrical launch because it's going to launch it's basically going to land at the same exact height that it started from. So this is basically a diagram. The first step is we want to break this up into its x and y paths, draw them and then draw the points of interest. So on the x-axis, we would just go straight from here to here passing the maximum height, which is always going to be a point of interest right there. So on the y-axis, we're just going to go up and then come straight back down again. So this is going to be our initial point. Then this point here where our maximum height is going to be b. And then finally we're going to hit the ground again at point c. So these are our points of interest here and the y-axis would be going from a up to b and then back down to c again. That's this first step. The second step is we're going to determine the target variable. What are we looking for in this problem? We're looking for the gravitational acceleration. So remember that that variable is little g. So where do we see little g in our equations? Well, in our y-axis equations, little g is always inside of this a y term. So we're really looking for when you were asking for the gravitational acceleration, because it says magnitude, all these numbers are going to be positive here, but we're actually really looking for a y here. So that's going to be our target variable. So the next step is what interval are we going to look at? Well, here's where you actually have some options because in symmetrical launches, remember we have a bunch of special properties about symmetry. We can use a to b or you can use a to b to c. And so we can always just choose one interval and if it doesn't work, then we'll choose a different one. So let's go ahead and choose the interval from a to b. So if we choose the interval from a to b and we're looking for a y-axis variable, then we're going to have to go ahead and list all of our variables out. Whoops. So in the y-axis, I've got a y which before we just said was equal to negative g. However, because we're not on the earth, we can't say it's negative 9.8. That's the whole point of this problem is the gravitational acceleration of this different planet here is not going to be 9.8. So don't make that mistake. So always be on the lookout for these kinds of problems. We can always assume 9.8 if we're on the earth. Okay. So this is actually where we're going to be going to be looking for. So what are our other variables? Our v naught y is v a y. V, our final y is going to be v b y. And we've got delta y from a to b, and then we've got t from a to b. Okay. So what about our initial velocity in the y-axis? Well, let's see. We've got a launch velocity. This is v naught or v a. We know this is equal to 10. We also have the angle is 37 degrees. So we can actually break this up into its x and y components, v a x and then v, sorry, v a x and v a y. And let's see. Our v a x is just going to be, the v x throughout the whole entire motion. That's going to be 10 times the cosine of 37 which is 8 and then our v a y is just going to be 10 times the sine of 37 and that's going to be 6. So we know what our initial velocity is. In the y-axis, That's just 6. What about the final velocity in this interval? So basically we're only just looking at this interval right here from a to b. So we start off with, 6 meters per second in the y-axis, and then what do we end with when we get to point b? Well, remember that in vertical motion or projectile motion, the reason that this maximum height is very useful is because VBY is equal to 0. We know that the velocity, once it gets up to its peak, is going to be momentarily 0 there. So we know this is going to be 0. And so what about delta y from a to b? Well, that's going to be the distance, the vertical distance from a to b and we don't know that. What about time? So do we know time? We also don't know the time either. So it looks like we're kind of stuck here. We don't know the time and we don't know delta y from a to b. Of these two variables, the one that I could solve for by going to the other equation is going to be the time. Remember where we're stuck in the y-axis, we go to the x-axis and I can't solve for delta y from a to b in the x-axis. So instead, what I can do is I can solve for time. So let's go ahead and do that. Then I'm going to go to the x-axis and I want t a to b. So I'm going to use the equation delta x from a to b is equal to v x times t from a to b. Now what happens is what is the delta x from a to b? Well, the only delta x, the only horizontal displacement that I know is I know that the horizontal displacement from a to c is equal to 32 meters. So I know this whole entire thing here. So let me see if I can write this. This whole entire thing is 32 meters. But from symmetry, one of the things we can do here is we can basically cut this projectile motion basically in half, then that means if the whole entire horizontal displacement is 32, then each one of these pieces here is 16 meters. From a to b, it's 16 meters and from b to c, it's 16 meters. So it means that delta x from a to b is 16. My x velocity here is 8 and T A B, is my unknown variable. So that means if I just go ahead and solve for this, my T TAB is equal to 2 seconds. So now I can actually plug this back in here. I know that's 2 seconds now. And so now I have 3 out of 5 variables and I can go ahead and pick the equation that ignores delta y. So if I go ahead and do that, that's just going to be equation number 1 and this says that the final velocity V B Y is equal to my initial velocity V A y plus A y times T from A to B. I know this is going to be 0, freebie 0. I know this is going to be, 6 plus and then I've got a y times 2 over here. So if you go ahead and solve for this, what you're going to get is that a y, which is equal to negative g, is just negative 3 meters per second squared. So what does that mean? It means that the magnitude of the gravitational acceleration is 3 meters per second squared. And so if we go to our answer choices, that is answer choice c. Alright, guys. That's it for this one.
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5. Projectile Motion
Symmetrical Launch
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