Negative (Downward) Launch - Video Tutorials & Practice Problems

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1

concept

Solving Downward Launch Problems

Video duration:

9m

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Hey, guys. So in this video, I'm gonna show you how to solve problems in which objects are launched downwards. So it's just another kind of project on motions. We're going to stick to the same system, the same steps and equations we've been using so far. So let's check it out. We've got a rock that is thrown at some speed and angle, and we're gonna calculate the height of the building in this first part. Remember, Before we get into any equations or anything like that, we're just gonna have to draw everything else. So once this rock is thrown, it's gonna be under the influence of gravity. Where to draw the past in the X and Y axis could only move along the X It would go from here, here along the y axis. It would go from here to here now one of the points of interest. Remember, the initial is always your first point of interest over here at this point A and that anything else that happens in between. But really, it just goes from the the table or the building or whatever, and it goes down to the ground like this. So that's our second point of interest. So that's the first step. So now we're gonna do is determined to target variable here. So that's the second step. And what target variable is that? That's the height of the building. So we're looking for the height, Someone to call that h And what is that actually represents? Well, this height of the building here is gonna be this variable or this, uh, this sort of vertical distance here in my diagram. So if I'm going from A to B in the Y axis, the height of the building is actually the same exact thing as the vertical displacement. That's gonna be Delta. Why? I remember you always gonna look for one of these variables that are in your, um, equations and one of these equations over here. So the only one that corresponds to a vertical displacement is gonna be my delta y. So that's gonna be my target variable. So now that moves to the left that moves us to the third step, which is we're gonna determine the interval and our, um, equation. So which interval? We're looking at here. Well, on the y axis, we're going from a down to the ground at point B. So that's gonna be our interval. So it's gonna be our interval from A to B. So we're looking for Delta Y from A to B. All right, so now we're gonna do is figure out the, um, equations, but because we're looking for a y axis variable, we're gonna start with the Y axis equations and variables. So we know that a wise negative g which is negative 9.8 initial velocity is gonna be V A y. Final velocity is gonna be v b y. They were looking for Delta y from A to B and then ta to be So we're looking for Delta Y from A to B, so we're gonna need three out of five variables. So our initial velocity in the Y axis over here. Can we figure that out? We have the initial velocity. That's Vienna is just equal to five, and we do. The angle here is 37 degrees, but because it's below the X axis, this is actually gonna be a negative 37 degrees. Because we have the magnitude and direction we can figure out the components V not X and not y So this is just gonna be my V a X. That's gonna be my V. Not times the cosine of fada. So this is just gonna be five times the co sign of 37. But we're actually gonna use negative 37 it's gonna be four. If you do the same thing over here, there's gonna be V A. Y is gonna be five times the sign of negative 37 we're gonna get negative three. So this makes sense. Because if you were to take this velocity and break it up into its components, this is gonna be a positive for because this points downwards, this V not Why is gonna be negative? Three. That's actually what's unique about these problems is that whenever you launch an object downwards, what's special about these kinds of cases is that v not Why will always be negative. That's what you need to know about these downward launch problems. All right, so we got to the initial velocity in the Y axis. That's negative. Three. I don't know what the final velocity is. That's basically the velocity when it reaches the ground over here, the white components this is V B Y. I don't know what that is. I'm trying to figure out Delta Y and I also don't know what the time is. I'm not told nothing about the time that it takes to hit the ground. So it looks like I'm kind of stuck here because I only have two out of five variables, but I need three. So what happens is remember, if you ever get stuck in one axis, you can always move to the other axis and try to solve for another variable. This happens in the X or Y. If you get stuck in the why we can always just go to the X axis to try and solve for another variable. Now, we can't go to the X axis to solve for V B y, but we can go to the X axis to solve for the time. So that's what we're gonna do here. So we're gonna solve for the time in the X axis someone look for ta. Be Remember, there's only one equation to use. That's Delta X from A to B equals v x times ta to be. Now do we know we know what this, uh horizontal velocity is What about the horizontal displacement from A to B? We're told in the problem that it hits the ground 10 m away from the building. So this is my horizontal displacement from A to B. So Delta X A to B equals 10. So we should Do you know what this is? This is just 10. This equals four times ta be. Which means that TA B is equal to 2.5 seconds. So we actually do know what ta be is. Now we can plug it back into our y axis equations. So notice how. Here we have now three out of five variables so we can pick the equation that does not include are ignored variable. And so that's gonna be equation number three. That does not include the final velocity. So this is gonna be equation number three Delta y for me to be equals va y times t b plus one half a y t a b squared. Now we're just gonna plug and chug everything because we actually have all these numbers. This is negative. Three 2.5 plus one half times negative. 9.8 times 2.5 squared. If you go ahead and work this out, this is gonna be negative. 19.75 m. However, is this your final answer? Well, it depends on your professor, because you're looking for the height. Onda Heights usually aren't negative. So what we'll do is we'll take the absolute value of this displacement over here and say that the height is always gonna be a positive number. So the height of the building is 19.75 m. Alright, so that is actually our final answer. All right, let's move on to part B. We're gonna go through the same list of steps. Um, remember that we already have the past and X and y, so we don't need to do anything. What about the target Variable? Well, for part B, we're trying to figure out the magnitude and the direction of the velocity before hitting the ground. Remember, it hits the ground here at point B. So they're not asking for the components of the velocity V B Y or VB X. What they're actually asking for here is they're asking for the magnitude of the two dimensional vector which these components make up VB So we're looking for VB and they're looking for the angle fate of B. So if I'm looking for the magnitude of a vector and the angle, then I'm just gonna use my Pythagorean theorem and my tangent Inverse equations. So this is gonna be vb X squared plus V b y squared and theta B is gonna be the tangent in verse off the absolute value of V B Y over vb X. So if I can figure out both of those components, then I can figure out the magnitude and direction. So if you think about this the X component of the velocity I already know because remember, the X component is always gonna be the same number. It's gonna be the 4 m per second throughout the entire motion. What? I don't know, because I don't know the final velocity in the Y Axis V B Y. If I can figure that out, then I configure the magnitude and the direction. So I'm actually gonna do that over here. So v b y is equal to what? So this is a Y access variable. So the next step is I just need to figure out the interval. Remember, we're only just working with one interval in this problem. The interval from A to B. So we're looking for the final velocity once it finally hits the ground in the UAE path. So I actually have all of my y axis variables over here. I'm just gonna copy paste. Um, so I'm just gonna copy paste these over here, Okay? So the only thing that's different now is that I have I'm looking for V b y. And I actually have Delta y from A to B. That's just negative. 19.75. So if you think about this, actually have four out of my five variables, which means I could just use any equation to figure out the last one. My final velocity. The easiest equation to use in this situation is the first one because it's the least amount of unknowns. So I'm gonna use equation number one V. B Y is equal to the way v a y plus a y times t from A to B. So this is just gonna be V. B y equals negative three plus negative 9.8 times 2.5 and you're gonna get negative, 27.5 meters per second. Remember, this isn't your final answer, because we're gonna take this velocity and you have to put it back into your magnitude and direction. So for the magnitude, we're gonna get the square roots of four squared plus negative 27. squared and you get 27.8 and your angle theta B is gonna be the tangent inverse of your absolute value of negative 27 point 5/4. And what you're gonna get is you're gonna get 81.7 degrees. So we go to our diagram here. What this means is that the magnitude of VB is 27.8 m per second, but the angle, because it points below the X axis just like the initial angle does is gonna be negative. 81.7 degrees. That's important. Make sure you keep the negative sign there to indicate to your professor that's below the X axis. So these are your two answers choices right here. And that's it for this one. Let me know if you guys have any questions

2

Problem

Problem

Water pours from a spout at the end of a gutter with a speed of 3.5 m/s, where the spout is angled 45° downwards. The magnitude of the water's velocity when it hits the ground is 14 m/s. How high is the spout from the ground?

A

18.8 m above the ground

B

10.6 m above the ground

C

10.0 m above the ground

D

9.4 m above the ground

3

example

Cannon Firing Downward

Video duration:

7m

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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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