Guys, let's check out this problem here. So we've got a cannon that's firing a projectile with speed and angle; we're told that it's 49 degrees below the horizontal. What that means is that it's going to be a downward launch problem. So let's just go ahead and stick to the steps. First, we need to draw a diagram, draw our path in the x and y axes. So we've got a cannon that's fired, and then it's eventually going to move downwards and hit the ground. So our cannonball will look something like this. We're told that this initial velocity is going to be downwards. We know that the angle relative to the horizontal is negative 49 degrees, and then it's just going to take a downward parabolic path until it hits the ground. In the x axis, if you were to move along the x only, it would look like this; in the y axis, it would look like this. So where are points of interest? Well, it's always just the initial, and then what happens is it's just going to hit the ground later, that's the only other point of interest that we're told, so that's just point b. So that means our path in the x and y looks just like this. So that's the first step. What are we actually looking for? 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 what we're looking for here is the horizontal distance covered between points a and b. So this actually takes care of steps 2 and 3. We have the variable and also the interval. We're looking for the interval from a to b because we're looking for the distance between the point where it's fired, and then the point where it hits the ground. In the x-axis, we only have one equation, Δx from a to b = vx times t. So we're looking for Δx. I'm going to need my initial velocity in the x axis, which I can get because I have the magnitude, I have the v0 is 73, and I have the angle. So my v0x, which is vax, which is just vx throughout the whole thing, is going to be 73 times the cosine 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 in the y axis, vay, is going to be 73 times the sine of negative 49 with the negative sign, and you're going to get negative 55.1. The reason it's negative is that the initial velocity in the y axis points downwards. So that makes sense. So, I mean, we have these vx components. The one thing we don't have, though, is we don't have the time. We don't know how long it takes to get from a to b. So we're stuck here, and so we're going to need to go to the y axis to solve. We need my 3 out of 5 variables. The first one we always start with is ay, which we know is negative 9.8. Our v0y which is vay, we just figured out, is negative 55.1. The final velocity is going to be the velocity at point b, so that's going to be basically what's the y component of the velocity here, which we don't know. Then we have Δy from a to b, and then we have t 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're going to need one out of these variables, we just need only one more. So between vby and Δyab, which one do we know? Well, let's see. We're told that the fort is 300 meters above the ground, which means 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 going to be negative 300. So we've got our 3 out of 5 variables, 1, 2, 3. So we're going to 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 3. So that is going to be the equation that we use. So we're going to use equation number 3, which is Δy from a to b is equal to vayta+1/2 ay t2ab. Remember, I came over here looking for the times, that's what we're really solving for. So we have Δy from a to b, we have vay, and we have ay, so we just plug in everything. Right? So this is going to be negative 300 equals negative 55.1 times t from a to b, plus one half negative 9.8 times tab2. So these are all just numbers, and the one variable that remains here is tab. How do we solve for this? Well, if you look at this equation here, we can rearrange this because we've got some terms of tab and then we've also got some terms of tab2. I'm going to rearrange this equation. First of all, I'm going to do one half times negative 9.8, and I'm going to put this out in front. So I'm going to do negative 4.9. That's just what happens when you cut negative 9.8 in half. So this is going to be negative 4.9tab2, and then I'm going to do minus 55.1tab, that's this term over here. And then I'm going to move this term to the other side, this negative 300 to the other side, and it becomes positive 300, and that's going to equal 0. The reason I'm going to do this is that if you take a look now, this equation is kind of in the form a times t2, plus b times t, plus c, which is a constant equals 0. The reason this is useful for us is that we have these coefficients here, a, b, and c, which are all just these numbers that go before the t's, and the way we solve this kind of equation here is by using the quadratic formula. So we 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 t is going to have 2 solutions. It's going to be negative b remember you probably learned a song for this, plus or minus the square root of b2 minus 4ac all over 2a. So what happens is if you just plug in all of these numbers into this formula, then you're just going to get 2 solutions for t. I'm going to go ahead and work this out for you guys. So negative b. Well, b in this case is going to be negative 55.1. So the negative of a negative is going to be a positive 55.1, plus or minus the square roots of negative 55.12 minus 4 times a, which is negative 4.9. Keep track of the negative signs there, and this is going to be 300. Okay. So now we're going to do all this divided by 2 times negative 4.9. Okay. So if you go ahead and work this out in your calculator, the plus and minus case, then what you're going to get is you're going to get 2 solutions for time. You're going to get negative 15 and then you're going to get 4.0 seconds. Remember, because one 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 it doesn't make sense. Okay. So that means our time here is actually 4.0 seconds, which is great because remember we needed time to plug it back into our x equation to figure out what the Δx is, the horizontal displacement. So we actually do that down here. I'm going to say that Δxab is equal to vax, vx times tab. And so Δx is just going to equal the horizontal velocity, which is 47.9 times 4.0. And if you go ahead and work this out, what you're going to get is you're going to get 191.6, which can be rounded to 192 meters, 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. So, you know, hopefully, be on the lookout for that. Anyway, so that means our answer choice is going to be answer choice b. Let me know if you guys have any questions. Let's keep going.
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5. Projectile Motion
Negative (Downward) Launch
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