5. Projectile Motion

Negative (Downward) Launch

# Cannon Firing Downward

Patrick Ford

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Hey, guys, let's check out this problem here. So we've got a cannon that's firing a projectile toward the speed and the angle. We're told that it's 49 degrees below the horizontal. What that means is that it's gonna be a downward launch problem. So let's just go and stick to the steps. First we need to do is just draw a diagram draw are passing the X and y axis. So we've got a candidate's fired and then eventually going to move downwards and hit the ground. So our cannon is gonna look something like this, our cannonball. So we're told that this initial velocity is gonna be downwards. We know that the angle relative to the horizontal is negative 49 degrees, and then it's just gonna take a downward parabolic path until it hits the ground. So, in the X axis, if you were the only move along, the exit would look like this on the y axis. It would look like this. So we're points of interest. Well, it's always just gonna be the initial, and then what happens is it's just gonna hit the ground later. That's the only other point of interest that we're told. That's just point B. So that means our passing the X and Y just look like this. That's first step. So what, are we actually looking forward? That's the second step. What's the target? Variable. We're looking for the horizontal distance that the cannonball travels before it hits the ground. So we're actually looking for here is we're looking for the horizontal distance covered between points A and B, so that actually takes care of steps two and three. So we have the variable and we also have the interval. We're looking for the interval from A to B because we're looking for the distance between the point where, um, it's fired and then the point where hits the ground. Okay, so in the x axis, we only have one equation. Delta X from A to B equals V x times t. So we're looking for Delta X. I'm gonna need my initial velocity in the X axis, which I could get because I have the magnitude I have. The V not is 73 I have the angle. So my V not X, which is V a X, which is just the extra whole thing, is gonna be 73 times the co sign of negative 49 degrees. We go ahead and plug this into your calculator with the negative sign, then you're just going to get 47.9 and then your initial velocity and the Y Axis V A Y is gonna be 73 times the sign of negative 49 with the negative sign and you're gonna get negative 55 point. One of the reasons negative is because the initial velocity in the Y axis points downwards. So that makes sense. So, I mean, we have this Vieques components. The one thing we don't have, though, is we don't have the time. We don't have how long it takes to get from A to B. So we're stuck here, and so we're gonna need to go to the y axis to solve. So we need to buy three or five variables. So the first we always start with is a Y, which we always negative 9.8 RV, not y, which is V A. Y. We just figured out is negative. 55.1. The final velocity is gonna be the velocity at point B. So that's gonna be basically what's the why component of the velocity here, which we don't know. Then we have Delta y from A to B, and then we have tea from A to B. Remember, we came here because we were looking for time so that we can plug it back into this equation here. So we don't need one out of these variables which need only one more. So between V b. Y and Delta y Abe, which one do we know? Well, let's see. We're told that the fort is 300 m above the ground, which means that the horror that the vertical displacement from A to B as we're going along this path here this horizontal displacement we know is 300. But because we're going from top to bottom, then that means this is gonna be negative. 300. So we've got our three out of five variables 12 and three. So we're gonna pick an equation that ignores this final velocity here. So let's go ahead and do that. And that equation that ignores the final velocity is going to be equation number three. So that is gonna be the equation that we use So we're gonna use equation over three, which is Delta y from A to B is equal to V A Y t a b plus one half a Y t a b square. Remember who came over here looking for the times? That's where really where we saw them for So we have Delta y from A to B. We have v A Y, and we have a y. So we just plugging everything. Right, So this is gonna be negative. 300 equals negative, 55.1 times t from A to B plus one half negative, 9.8 times ta b squared. So these are all just numbers and one variable that just remains Here is t A B. So how do we solve for this? Well, if you look at this equation here, we can actually rearrange this because we've got some terms of ta be Then we've also got some terms of ta B squared. So what I'm gonna do is I'm gonna rearrange this equation. First of all, I'm gonna do one half times negative 9.8, and I'm gonna put this out in front, so I'm gonna I'm gonna do negative 4.9. That's just what happens when you cut negative 9.9 in half. Um, so this is gonna be negative 4.9 t a B squared, and then I'm gonna do minus 55.1. Ta be. That's this term over here. And then I'm gonna move this term to the other side. This negative 300 to the other side becomes positive. 300. That's gonna equals zero on. The reason I'm gonna do this is because if you take a look now, this equation is kind of a form a times T squared plus b times t plus C, which is a constant equal zero. And the reason this is really useful for us is because we have these coefficients here A, B and C, which are all just these numbers that go before the tease on the way that we solve this kind of equation here is by using the quadratic formula. So we're gonna use the quadratic formula sometimes to solve your projectile motion questions. And basically, the quadratic formula says that if you want to solve for t, your tea is gonna be is gonna have to solutions. It's gonna be negative. Be. Remember, you probably learned a song for this plus or minus the square root of B squared minus for a C all over to a So what happens is if you just plug in all of these numbers into this formula, then you're just gonna get to solutions. 40. So I'm gonna go ahead and work this out for you guys. So negative b will be in this case is gonna be negative 55.1. So the negative of a negative is gonna be a positive 55.1 plus or minus the square roots of negative 55.1 squared minus four times a, which is negative. 4.9. Keep tracking the negative signs there on. Don't lose track of them on. This is gonna be positive. 300. Okay, so now we're gonna dio all this divided by two times negative. 4.9. Okay, So if you actually just go ahead and work out this in your calculator, the plus and the minus case, then what? You're gonna get it. You're gonna get to solutions for time. You're gonna get negative 1.5 seconds, and then you're gonna get 4.0 seconds. Remember it because one of these equations, one of these solutions won't make sense. And this one doesn't make sense because it doesn't make sense to have a negative time, so doesn't make sense. Okay, so that means that our time here is actually 4.0, seconds, which is great, because remember, we needed the time to plug it back into our ex equation to figure out what the Delta X is the horizontal displacement. So we actually actually do that down here, I'm gonna say that Delta X A b is equal to V a x V x Chinese ta be. And so, Delta X, it's just gonna equal the horizontal velocity, which is 47.9 times 40 And if you go and work this out, what you're gonna get is you're gonna get 191.6, which could be rounded to 192 m. And so that is your answer for the horizontal displacement. So sometimes you may have to use the quadratic formula inside of these cases. Here s O, you know, hopefully be on the lookout for that anyway, eh? So what? That means that our answer choice is going to be the answer choice. Be so let me know. If you guys have any questions, let's keep going.

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