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Gravity on Planet X

Patrick Ford
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Hey, guys, let's check out this problem. We're playing catch on a far away planet. We've got some information about the launch speed and angle of a projectile and the distance that it travels, and we're gonna calculate the gravitational acceleration on this planet. So let's just stick to the steps. Basically, what I've got here is we've got a level ground. I'm told that I'm throwing a ball upwards with some velocity at degrees, and then it's gonna later return to the ground again. So that means when I draw out this trajectory, it's actually just going to be a symmetrical launch because it's gonna launch. It's basically gonna land the same things like height that it started from. So 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 in the X axis, we would just go straight from here to here, passing the maximum height, which is always gonna be a point of interest right there. So and in the Y axis, we're just gonna go up and then come straight back down again. So this is gonna be our initial points. Then this point here where a maximum height is, is gonna be be. And then finally, we're gonna hit the ground again at point C. So these air points of interest here on the Y axis will be going a from a up to be and then back down to see again. So that's this first step Second step is we're gonna 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 and 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 gonna be positive here, But we're actually really looking for a y here. So that's gonna be our target variable so that the next step is what interval we're gonna 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 be or you can use a to B to C s so you 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 Interval from A to B and we're looking for a y axis variable that we're gonna have to go ahead and list all of our variables out groups. 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. 98 That's the whole point of this problem is the gravitational acceleration off this different planet here is not going to be 9.8. So don't make that mistake s always be on the lookout for these kinds of problems. We could always assume 9.8 if we're on the earth. Okay, so this is actually where we're gonna be looking for. So what are other variables? RV Not y is va y V Our final Why is gonna be V B y? Anyway? You've got Delta y from a to B, and then you've got to tea 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 Knots 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 via Sorry via X and V A y. And let's see, R V A X is just gonna be, um, the VX throughout the whole entire motion. That's gonna be 10 times the co sign of 37 which is eight. And then our v A y is just going to be 10 times the sign off 37. And that's gonna be six. So you know what? Our initial velocity is in the y axis. That's just six. 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, uh, 6 m per second in the Y axis. And then what do we end with when we get to point beat will remember that in vertical motion or projectile motion. The reason that this maximum height is very useful is because VB Y is equal to zero. We know that the velocity, once it gets up to its peak, is gonna be momentarily zero there. So we know that this is gonna be zero. And so what about Delta y from A to B? Well, that's gonna be the distance, the vertical distance from A to B. And we don't know that, Um, 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 off these two variables, the one that I could solve for by going to the other equation is gonna be the time. Remember where we were stuck in the Y axis would 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 dio is I consult for time, So let's go ahead and do that. Then when you go to the X axis and I want t a to be so I'm gonna use the equation. Delta X from A to B is equal to v x times t from a to B. Um, now what happens is what is the Delta X from A to B well, the Onley Delta X The only horizontal displacement that I know is I know that the horizontal displacement from A to C is equal to 32 m. So I know this whole entire thing here. So let me see if I can write this. This whole entire thing is 32 m, but from symmetry when when things we can do here is we could 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 m from a to B. It's 16 m and from B to C at 16 m. So it means that Delta X from A to B is 16. My ex velocity here is eight and ta be eyes my unknown variable. So that means if I just go ahead and saw for this, my t A B is equal to two seconds. So now I can actually plug this back in here. I know. That's two seconds now. And so now I have three out of five variables, and I could go ahead and pick the equation that ignores Delta. Why? So if I go ahead and do that, that's just gonna be equation number one. And it 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 gonna be zero freebie y zero. I know this is gonna be six plus and then I've got a Y times two over here. So if you go ahead and solve for this, what you're gonna get is that a Y, which is equal to negative G is just negative. 3 m per second squared. So what does that mean? It means that the magnitude of the gravitational acceleration is 3 m per second squared. And so if you go to our answer choices, that is answer choice. See? Alright, guys, that's it for this one