Positive (Upward) Launch - Online Tutor, Practice Problems & Exam Prep

In non-symmetrical projectile motion, objects can be launched from different heights, affecting their trajectories. When solving these problems, sketch the paths in the x and y axes, identify target variables, and choose appropriate intervals. For example, when calculating the vertical component of velocity just before impact, use the equation $v^d=v^a+at$. Understanding these principles simplifies calculations and enhances comprehension of motion dynamics.

Solving Non-Symmetrical Upward Launch Problems

Hey, guys. So in the last couple of videos, we saw how to solve problems in which an object is launched upwards and then returns to the same height from which it was launched. This was called symmetrical launches, but this won't always happen. You might see some problems where objects are launched and they return to different heights. There are really just two options. You could land at a higher height or a lower height. And so, in general, if an object is launched upwards and it lands at a higher or lower height, then the motion is going to be not symmetrical. So that's what I want to show you in this video, how to solve non-symmetrical upward launch problems. We're really just going to use the same steps and equations as before. There's just a couple of new things here. So what would this look like if you landed at a higher height, so this trajectory would be like this. It would go up and then before coming back down to the original height, it would hit its target or a wall or a building or something like that. And if you were launching at a lower height, I like to call these launch from height problems, then it would look like this. It would go up, reach its peak and then come back down again and then hit the ground. Alright?

So before we get into this next bullet point, I actually just want to go ahead and start the problem because we're just going to use the same exact system to solve them. So we've got this potato. We're going to fire it from a potato launcher while we're on top of a cliff, which is 20 meters high. And so before we go ahead and draw anything and start working out the variables and equations, we're just going to solve the first step. We're going to draw the paths in x and y. We've got this object here, and it's going to go up and down like this. If we could only move along the X-axis, it would look like this, and the Y-axis would go up to the peak and then come back down again towards the bottom. So what are our points of interest? Remember, we're looking for things like the initial and the final and the maximum height. So this is our initial, and then we've got the maximum height over here, and we've got the final which is where we hit the ground, but there's one other point of interest which happens in between. So what happens is this object is going to go come back around again. It's going to pass the heights from which it was initially launched and then continue further. So this is going to be b. We'll call this c and then d. So what happens is, if you're ever launching and landing at a lower height, then what happens is part of the motion from a to c is going to be symmetrical. So we see how from a to c, this is going to be just like a symmetrical launch problem. What's unique about these kinds of problems is that the object also continues to drop further in this interval from c to d. Alright?

Again, we're going to get back to the second bullet point right here. Let's just get right to the problem. So we've got the paths in x and y. Now we're going to determine the target variable. What we are actually looking for here is the vertical component of the potato's velocity just before it hits the ground. So that's not going to be point D. Remember, there are two components of the velocity. There's going to be the x component, which I'll call v_dx, which is just v_x because it never changes, and then we're looking for v_dy. So this is what we are looking for here. Remember these two components combine to produce a two-dimensional vector v_d, but that's not what we're looking for. We're just looking for v_dy. So we're looking for that target variable here.

Now, the next big question is: what interval are we going to use? Remember, we have a bunch of different points of interest now. There are many different options as to the interval that we could choose. There are a couple of options here, and depending on which one you take, you can still get the right answer. But there are a couple of options that are easier and harder. One thing you might be thinking of is if you have the initial velocity, which is v_a, which is just equal to 30, then what you could use is a symmetry argument. You could say, well, if v_a is 30, then I know that v_c is also going to be 30, except it's just going to be pointing in the opposite direction. So I could use my components v_cx, which is v_x, and then whatever v_cy is is just going to be the negative of v_ay. So you could use this interval just from c to d, and you absolutely could solve it that way. But I'm going to warn you that there's going to be a bit more work, and you're going to have to solve a bunch of other things. The easiest choice for these kinds of problems and, in general, what you should try to do if you're ever landing at a lower height is b, which is the maximum height. You're always going to want to try to include this in your equations because one thing that we know is that the y velocity when it's at the peak is equal to 0, and we're going to see how this simplifies our equations and our variables. Alright? So we're going to be looking at the interval from b to d. So we're looking for the y-axis, so I need all my y variables. So I've got my acceleration in y, which is always negative 9.8, the initial velocity. Well, the initial velocity in this interval from b to d is actually going to be v_by and that's 0. This is why this interval is really easy to use and you should always try to use it because we've already unlocked. We've already got two out of the five variables that we need. We just need one more, and then we'll be able to solve our equations. So the final velocity is going to be at point d, which is exactly what we're looking for. And then we've got Δy from b to d and then t from b to d. So we just need one of these other variables here.

Let's take a look at Δy from b to d. So what does that mean? That would actually represent the vertical displacement from point b all the way down to the ground like this. So the potato is launched from a 20-meter-high cliff. So is that the Δy from b to d? Well, no, because that's only going to be this piece right here. This vertical distance is the 20, not this whole entire thing over here. However, what we're also told from the problem is that the potato reaches its maximum height of 49.4 meters above the ground. So this is going to be our Δy from b to d. However, because this points downward, it's going to pick up a negative sign. So that's Δy, which is going to be a negative 49.4. And then we don't know the time. We don't know how long it takes to go from the peak back down to the ground again, but that's okay because we've already got 3 out of 5 variables here. Alright? So we're just going to use the one that ignores time, that's equation number 2. So the final velocity which is v_dy squared is going to be the initial velocity v_by squared plus 2a_yΔy from b to d. Okay?

So we already know that the y velocity v_by is going to be 0. And so v_dy squared is just going to be 2 times negative 9.8 times negative 49.4. And so if you go ahead and work this out, we're going to get that v_dy is equal to the square root of 960, which is just going to be approximately 31.1. However, there are two answers here. We could use 31.1, which is going to be a positive 31.1, which means that we're getting a velocity that points up, or we could get a negative 31.1 because remember we're taking the square root of a number here and we're solving for velocity. That would correspond to a velocity that points downwards. So which one of these answers is correct? Well, if we take a look at v_dy, it points downwards. So that means that the positive answer is not the correct one. Instead, we're going to use v_dy is equal to negative 31.1 meters per second. That's the vertical component of the velocity right before it hits the ground. That's it for this one, guys. Let me know if you have any questions.

You throw a rock off the top of a tall building at an upward angle of 15°. At t=3 s, the rock's horizontal displacement from you is 52m. How high does the rock get above the top of the building?

• A. 1.1 m
• B. 4.6 m
• C. 30 m

Height of a Roof

Hey, guys. Let's check out this problem here. We've got a child who is throwing a ball up to a rooftop and we're going to figure out how high the roof is. So let's go ahead and draw a diagram. I've got the ground level like this, and then the rooftop like this. We've got this ball that's being thrown at some angle like this, and it's going to go up, and what else are we told? We're told that the ball is going to reach its maximum height directly above the edge of the roof. So when we go to draw our diagrams, it's actually really important that we draw it the correct way. This is saying here that the maximum height of the trajectory is going to be right above the edge of the roof, and then it's going to fall back down, and it's going to fall back down, and we know that this is going to be 3 meters away from the edge of the roof. So let's just go ahead and stick to the steps. We're going to draw the x and y paths and the points of interest and then start filling out everything we know about the problem. So in the x-axis, we're going here, and in the y-axis, we're going up to the maximum height and then back down to this point right here. So our points of interest are a, the initial, b, which is the maximum height, and then finally when it hits the rooftop which is at point c. So in our x, it's going to look like this, and in the y, it's going to look like this. Alright. So that's our path in the x and y. And let's see. What else do we know? We know the initial velocity here is 13. We know the angle is 67.4 degrees. Alright. It's basically all we know here. So let's go ahead and move on to the second step. We're going to determine the target variable. What are we looking for? We're looking for how high the roof is. So if we look at our diagram here, what that distance would represent is it would actually represent this vertical displacement here. And if you look at it, that's basically from the ground where the child launches the ball from, and then to the place where it hits the roof. So in the y-axis, what we're really looking for is we're looking for the distance between a and c. So this is delta y from a to c here. So that's actually going to be our target variable. Delta y from a to c is going to be the height of the roof, which actually brings us to the third step, the interval and which equations we're going to use. Well, if we're looking for a to c, then one interval we can choose is the interval from a to c, but what happens is if you just go ahead and write out all these variables, you're going to end up figuring out that you can't actually solve this. So I'm just going to warn you right there, if you try to do it, you're going to get stuck. So we're going to need a different kind of approach for solving delta y from a to c. Now one way you could do this is you can actually break up the motion into several parts. If you wanted the delta y from a to c, that would actually just be the distance from a to b, right? So this would be delta y from a to b plus the downward displacement from b to c. You could basically, if you could figure out both of these vertical displacements, then you would be able to add them together, and that would represent the difference between them, delta y from a to c. So this is actually going to be the approach that we could take for this problem. Delta y from a to c is delta y from a to b plus delta y from b to c.

The reason this is useful is because we know a lot of information about this point b here. We're going to be able to use 2 out of 5 variables that we already know. So basically, I just have to figure out both of these vertical displacements, add them together, and then that will be my final answer. So that brings us down to the next step. So we can't use the interval from a to c, but if we want to figure out delta y from a to b, then I'm just going to use the interval from a to b instead. So I've got my Ay which is -9.8. We've got my v₀y, which is v_ay, which is, let's see. Actually, let me go ahead and finish out the rest of the variables. V_y is v_by, and then we've got delta y from a to b, and then we've got t from a to b.

So what's the time? Alright. So let's go through each of these variables, v_ay. So if we have the magnitude which is 13 and the direction 67.4, so we can figure out v_0x which is v_ax which is just 13 times the cosine of 67.4. And if you do this, you're going to get 5. If you do v_0y which is v_ay, this is going to be 13 times the sine 67.4 and you'll get 12. So we know this is 12 over here. What about v_by, the final velocity? Well, remember that's an important point in our projectile motion problems because we know that v_by, basically your velocity at the top here is equal to 0. Once you hit the top of your peak, your velocity is instantaneously 0. So it's momentarily 0 for that little point there, so that's 0. So that means that we have 3 out of 5 variables, and we're looking for this Delta Y from A to B, which means I can just go ahead and pick the equation that ignores time. So that's what we're going to do here. So that's equation number 2, which says that the final velocity v_by squared is v_ay squared plus 2a_y times delta y from a to b, right? We know this is equal to 0. That's why it's important and useful. We know this is 12 squared plus 2 times -9.8 times Delta y from a to b. If you go ahead and work this out, you're going to get Delta y from a to b is equal to 7.35 meters. So this is only just part of the piece. Remember this distance here is just only one of the pieces of our equation, so we need 7.35 over here, and then we just need to figure out delta y from b to c. Right?

So what we're going to do is now that we're done with the A to B interval, we can just move on to the next one. So we can say in the B to C interval, I'm just going to write out the same variables again. Right? So I've got -9.8. Now my initial velocity from the B to C interval is going to be v_by. That's my initial velocity. We know that's 0. What about v_y, the final velocity? That's going to be v_cy, and we don't know what that is, and then delta y from b to c, which is what we're trying to find. Remember, that's going to be what we plug in over here, and we should expect it to be negative, and then t from b to c. Alright. So now it looks like we have 2 out of 5 variables, but I'm going to need either the final velocity or I'm going to need time. So remember, whenever I get stuck in the y-axis, I'm just always going to go over to the x-axis. So the easiest one to solve in the x-axis is going to be the time. So I'm just going to go over here, and I'm going to look for TBC, then I'm still just going to look at the BC interval here. I'm just going to write out my equation, Delta XBC equals V_x times TBC. So do we have both of these other variables? Remember, we're looking for time, so do we have v_x? Of course, we do. So because we just figured that out in the last part, we know that's equal to 5. What about delta x from b to c? Remember that was the horizontal displacement from b to c, that would actually represent this distance right here from b to c. We actually know what that is because remember that we're told the ball lands 3 meters from the edge of the roof, and we know that the distance from the edge to the roof over here is going to be the horizontal displacement from b to c. So that's delta x. So this is actually what delta x from b to c is. You should write that down. So let's move on. So now we actually have both of these variables. This is just 3, and then we have 5 times TBC, which means that TBC is equal to 0.6. So now that we have this 0.6, then we can go back to our equations in the y-axis, and we have 3 out of 5 variables, 123. So we're going to go ahead and pick the one that ignores the final velocity, so we get delta y from b to c. I'm going to use equation number 3 this time. So delta y from b to c is v_y as the initial velocity times t_vc which we know is 0 because remember this is just equal to 0, plus 1 half times -9.8 times 0.6 squared. Right? So I'm just going to go ahead and fill those in, and what you get is that delta y from b to c is equal to -1.76. So this is the vertical displacement, and it totally makes sense that we got a negative number because remember, the vertical displacement between these two points because it points downwards should be a negative number. So we get -1.76 here. So when you add these two things together, you're going to get 5.6 meters. And that is your final answers. That is the height of the building. So that's answer choice c.

Alright guys. So we can see here whenever you get stuck using one interval, you can always basically just break it up into a combination of intervals and then go ahead and figure out what your target variable is using that method. Alright? So let me know if you have any questions, and I'll see you in the next one.

Catapult Siege

Hey, guys. Let's work out this problem together. So we've got this catapult that's launching a stone. It's going to go up and then it's going to hit a castle wall. We're trying to figure out the direction of the stone's velocity. So let's just go ahead and stick to the steps. We're going to draw the paths in the x and y and then label our points of interest. When you go to do this step, you might actually realize that there are two different ways to draw the path of the stone. One possibility is that you could hit the castle wall while still on the way upwards in which the final velocity is going to be upwards like this. Or the other situation is that the catapult might be launched upwards, reach its maximum height, and then actually hit the castle wall on the way back down. And we don't know from the situation or from the information that's given in the problem which of these situations is the actual correct one. But it turns out that it doesn't really matter for our purposes. So we're going to draw the path in the x and y. So the x path is going to look the same for both of the, it's just that one of them is going to be a little bit shorter, but the y path might look different. The y path for this one is just going to go straight up, and for this one, it's going to go up and then come back down again. So when we label our points of interest, there are actually two possibilities. You could just be going from a to b, in which case your interval is going to look like this, a to b, or you could be going from a to b and then back down to c in which you hit the castle wall. So you're going to be looking like this, a to b to c. So because we actually don't really know, you should always just try to draw the passing the x and y. And if you can't tell, you can't actually figure out which one it is, it doesn't really matter because we can actually just, all we're looking for here is we're just looking for the point which it's launched to the point where it hits the wall. So I'm going to call this interval even though I don't know which one a to c. So what we're really looking for here is we're looking for the angle that this velocity makes with the horizontal. And so remember, there are two components. We've got v_cx and v_cy. And really, what happens is because we know the x velocity is never going to change, what's really going to affect this is the y velocity. So let's go ahead and go to the second step here, which is our target variable. What are we looking for? We're looking for the direction which is θ, and we're looking for the direction at point c. So that's going to be this variable over here. Alright. So because θ_c is actually just a vector equation, it's just this, tangent inverse over here, I'm going to start with that. So we have tangent inverse of and we need v_cy over v_cx. So v_cx is the easy one because remember that the x component of the velocity never changes. So what I can do here is say that my v_cx is just going to be v_x throughout the whole entire motion. And that's actually just the initial velocity times the cosine of θ, right, which I know that's just 50 times the cosine of 56, and I get 28. So that's never going to change. It's always going to be 28 throughout the entire motion there. V_cy is where it gets a little tricky because I don't know what this v_cy is, but I'm going to have to go find it. I'm going to have to pick an interval and then solve it using some UAM equations. So that's the third step. So our interval that we said we're going to use is just the interval from a to c. So even though we don't know what the actual trajectory is, we don't know if it goes up or comes up and back down again, it doesn't really matter because all we're really looking for is the point where it's launched and the point where it hits the castle wall. So let's just go ahead and list out all of our variables here. In the y-axis I've got a_y is -9.8. I've got v_0y, which is v_ay, which I can actually figure out here. My v_ay is just going to be, or rather my v_{initial y} is going to be v_ay, and that's v₀ times the sine θ. So that's just 50 times the sine of 56, and that is 41.5. Man. So 41.5. Okay. So we know this is 41.5. And then our v_{final y} is going to be v_cy. This is actually what we're trying to look for. And then see, we've got Δy from a to c and then t from a to c. We do know that it takes 6 seconds for the stone to hit the castle wall, so that's t_ac. So that means that we have three to five variables, and we pick the equation that ignores my Δy. So this equation here is actually going to be equation number 1, which says that the final velocity, v_cy, is v_ay + a_y times t_ac. So v_cy equals our initial velocity in the y-axis, 41.5 plus -9.8 times how long it was in the air, 6 seconds. If you go ahead and work this out, you're going to get -17.3 meters per second. So this is our y components of our velocity. Now remember, this is not our final answer because what we need to do is we need to take this number and we need to plug it back into this equation over here, to get our tangent inverse. So what we're going to do here is θ_c equals the tangent inverse of we've got -17.3 divided by 28. That's our x components. And if you go ahead and work this out, what you're going to get is you're going to get -31.7 degrees. So which one of our two situations did it end up being? Well, because the y component ends up being negative, it's actually going to be this situation over here. This was the correct one. And so this angle over here, θ_c, because it's negative, means it's being below the horizontal. So your θ_c is -31.7 degrees, and that is your final answer. Alright. So that means that brings us to answer choice A. And let me know if you guys have any questions about that one.

Using Single Intervals in Positive Launch Problems

Hey, guys. So by now, we've gone through all the steps to solve projectile motion problems. For any given problem, we're just going to sketch the trajectory and draw its paths in the x and y. And once we figure out our target variable, we're going to pick our intervals and then start using our equations. And what we've seen in projectile motion is that often, you can choose different intervals and still get to the right answer. For example, imagine if I had this upward launch here to some lower height at point d and I wanted to calculate the total amount of time from initial to final. Well, actually, I have a couple of different paths that I can choose from, a couple of different ways I can solve for that. First, I can try to go from a to b, calculate that time, so delta t from a to b, and then add it to the time that it takes to go from b to d. So just adding those two smaller times there. But another option is I can also just try to calculate the time from a to c because I know that it is symmetrical and I can try to use some of the symmetry points there. So that's delta t from a to c and then add it to delta t from c down to d. So if I had enough information, I could actually solve and get the same time and same answer using both of these different approaches. But often, the easiest approach to solving these problems is just by using a single interval going from a all the way down to d. You can absolutely do that. And in fact, that's going to be your best bet on solving these kinds of problems. Usually, you should try as best as possible to solve these problems using a single interval like from a to d because it's going to make your life a whole lot easier. Let me show you how this works in this example down here. So we've got a cannon and we've got the initial speed and the angle and we're going to calculate the vertical component of the velocity at the point where it hits the ground. So let's just go ahead and go through the steps. First step is we're going to draw the paths in the x and y. So we're going to go ahead and have paths in the x and y. That's point b, c, and d and the y axis it goes up and then back down again. So that's b, c, and d. And now, that's the first step. The second step is we're going to figure out the target variable. So in this first part here we're going to calculate the vertical component of the velocity at the point where it hits the ground. That's point d. So remember at any point there are two components of the velocity. There's vdx, there's the x component and the y components, that's Vdy. And both of these combine to form a two-dimensional vector using the Pythagorean theorem, Vd. So which one of these variables are we looking for? We're looking for the vertical component that is Vdy. So that's our target variable. So now it just comes to what equation or what interval are we going to use? Well, remember in these problems, we're going to try as best as we can to use the interval from a to d if we have enough information. If we get stuck and it turns out we don't, we can always just come back and then pick some different intervals. Alright. So I'm looking for a y-axis variable. So that means I'm just going to use, I'm going to list out all of my y variables. Right? So this is going to be ay which is negative 9.8 and that's regardless of the interval that we're using. Remember ay or, is always always negative g. It's negative 9.8 no matter which interval you're going to choose. So our initial velocity is our velocity at point a. Our final velocity is vdy. That's actually what we're looking for. And then we've got delta y from a to d and then we've got t from a to d. So let's start knocking down these variables. So I'm trying to figure out vay and that is, I can figure that out as long as I know the initial velocity, which is va, which is 100, and the angle which is 30 degrees. I have both of those magnitude and angle so I can calculate the components. My v not x which is vax which is just vx throughout the entire thing, the entire motion will never change, is 100 times the cosine of 30 and that's 86.6. That number is never going to change throughout the whole motion. The y velocity on the other hand will change. That's vay and that will be 100 times the sine of 30 and that's going to be 50. So that's our variable there. We have VAY is equal to 50. So which other ones which other variables can we solve for? We've got delta y from a to d and then t from a to d. Well, the only other thing that we're told in the problem is that this cannon is fired from a 40 meter cliff. So that what what that means is that this height over here from the cliff down to the ground is 40. And if you think about it, this is actually just the vertical displacement from a down to d. Forget about what happens in the middle. So your interval is going from a all the way down to d, but all you really care about is the initial and final. And so because your displacement in the vertical is just going to be 40 downwards again, forget about what happens in between. It doesn't matter that it goes up and then comes back down again. It's going to be negative 40. So that's negative 40 over here and so, therefore, we have our three out of five variables. I've got 1, 2, and 3 and I'm going to pick the equation that ignores time. So that's equation number 2. So I've got vdy squared equals vay squared plus 2 ay times delta y from a to b. I have all those numbers and I'm just going ahead and plug. So my vd is going to be the square root of, and then I have 50 squared plus 2 times negative 9.8 times negative 40. And if you go ahead and work this out, what you're going to get is you're going to get plus 57.3 and also minus 57.3. Because remember that whenever you take the square root of a number like the square root of 9, there are two answers positive and negative 3. Both of those things when you square them will both equal 9. So, which one of these makes sense? Well, a positive 57.3 meaning would mean that the velocity is going up. And a negative 57.3 means that it would go down. So obviously at point d, our velocity component is going to be downwards. So, therefore, this is not the right answer and this negative 57.3 degrees is our right answer. Alright. Let's move on. So this point part b now, now what are we looking for? We're looking for the total time, t from a to d. So we go through the steps again. We've already have our past in the x and y. We've already determined the target variable and the interval. We're still just going to try to use a to d. In fact, ta to d was the variable that we initially ignored in this first part. But now that we know vdy, now we know this is just negative 57.3, Now we actually have 4 out of the 5 variables. And so we can actually use any one of the equations to solve for t. Any one of these equations will be totally fine for solving the time. So what we're going to do is we're going to use equation number 1 because it has the least amount of terms, it's the simplest equation to use. So this just says that vdy here in our same interval, a to d, is just going to be vay plus, and this is going to be, negative Oops. Plus ay times ta to d. So we've got negative 57.3 equals 50 plus negative 9.8 times ta to d. If we go ahead and move this over, we're going to get negative 107.3 equals negative 9.8 times ta to d, and so we're going to get negative 107.3 divided by negative 9.8. The negatives will cancel and then you'll just get t from a to d which is equal to 10.9 seconds. So notice how we were able to solve all of these problems here, both of these parts, just by using a single interval. It's going to make your whole life a lot easier if you can try to do this as best, as much as you can. Anyway, that's it for this one guys. Let me know if you have any questions.

A ball is thrown from the top of a 50-m-tall building with a speed of 40 m/s at an angle of 37° above the horizontal. How far horizontally does the ball travel before hitting the ground?

• A. 101.2 m
• B. 50.6 m
• C. 207 m
• D. 414 m

Example 3

Everyone. Welcome back. So, in this practice problem here, it's kind of complicated, but we've got a ball that's thrown from a building at 37 degrees above the horizontal. And then basically what happens is later on, it's going to hit another building that's 24 meters away at a lower height, and it's going to break the window. So let's go ahead and just draw a quick sketch of what's going on here. Basically, we've got a building, the ball that's launched up like this, but then it lands lower and breaks a window on another building that's across the way somewhere. Alright? We want to figure out how high above the ground is the window, and we're going to round it to the nearest whole number. So here's what's going on. Right? We've got the height of this building, so in other words, this distance over here. I'll call this h_b for building, and that's equal to 50 meters. What we want to do is figure out not delta y, but we want to figure out the height of the window. So h_window, which is basically the height measured from the ground. How do we actually go about doing that? Well, this is going to be a projectile motion question, and the idea here is that we have a positive launch, and we're going to try to solve this all in one interval. So let's go ahead and stick to the steps. We're going to draw our paths in the x and the y and just label our points of interest. So in the x direction, it's basically just going to go like this. It's going to go straight across like that. In the y-axis, it's going to go up, and then it's going to come back down again. And what we know is that it's going to land at a lower height, so it's going to look something like this. So here are points of interest. We've got a, we've got b, which is the peak, c, which is where it's symmetrical, and then d, which is where it hits the window. Alright? And the and the y direction is going to look like this, a, b, c, d. In the x direction, it's going to look like this, a, b, c, and then d over here. Alright? So let's get started.

So, basically, what we know here is that the height of the window, how do we actually go about solving for that? How are we going to use a projectile motion or kinematics equation to solve that? Well, we know here that the height of the window of the building that it starts from is 50. But then what happens is that as the build as the ball flies through the air, it goes up, and then it comes back down again before it hits. So, basically, what happens here is it's going to have some displacement in the y direction. That's going to be delta y. So basically, what happens is the height of this window is just going to be the height of the building that it starts from, which is the 50 plus delta y over here. Alright? But now we're going to have to look at the target variable and figure out what interval we're going to use. If you look at this, what happens is this is just delta y. This is actually delta y throughout the entire motion. This is delta y from a all the way to d. It goes back it goes up again, goes to b, comes back down through c, and then finally goes and hits over here at d. So this is really delta y from a to d. That's going to be our target variable, so delta y from a to d. So that means that we're going to choose the interval from a to d because, usually, we're going to try to figure out if we can solve all of this just using one interval instead of breaking it up. Alright? So we're going to have to label our variables in the y-axis here from a to d. So I'm going to start with delta y from a to d. That's our target variable, so we're looking for that. We need our other 4 variables. So, for example, the initial velocity, that's v_a initial. So sorry. Let's do v initial, which is actually going to be v_a, right, in the y direction. We also have v final, which is going to be v__dy. We have the acceleration in the y axis, which is always negative g, which is negative 9.8. And then we've got time. That's delta t from a to d. Now, what about that, that time? We're actually told that this projectile flies through the air, and 3 seconds later is where it hits the window. That means that our delta t is actually equal to 3. We know what that is. So, so far, we've got 2 out of these 3, out of these 5 variables. We need 3. So that means we're going to need one of these other 3 over here. Alright? This is our target variable, so we really just need one of these, the initial or the final velocity. The initial velocity is always a little bit easier to solve for because we do have some information about it. For example, we know that the angle is 37 degrees above the horizontal. So in other words, this angle over here, we know that this angle here is 37. Unfortunately, though, we actually don't know the magnitude of the initial velocity. Alright? So what we know here is that this v_a_y is actually going to be v_a sine of 37. The problem is we actually don't know what this v_a is. Alright? So we're going to need one of these 2 to solve. And because we already have this equation set up, we're going to have to figure out this VA. How do we do that? Well, what happens is when we get stuck in the y axis, we're going to have to go and look at the x axis in order to solve for this. Alright? So let's go look at the x axis and see what we can solve here.

There's only one equation to look at in the x-axis, and it's basically just that delta x is equal to v_x times t. However, what happens is we're going to use the same exact interval. So in other words, the delta x from a to d is equal to v__ax times delta t from a to d. Alright. If you look at the x axis, here's what's going on. The ball flies through the air, and it basically just goes all the way across like this. And we actually know this distance over here, which actually is our delta x from a to d, it's basically the horizontal distance the ball travels, is equal to 24. We also know again that the time is just equal to 3 seconds. So, basically, what we know here is that this is 24, which equals v__ax times 3 seconds. And we can use this to solve. Basically, just divide by 3. That cancels. This is going to be divided by 3. In other words, we're going to get a is equal to v_a_x. And we know from our vector decomposition that this is actually going to be v_a times the cosine of beta. Alright? In other words, it's going to be v_a times the cosine of 37. The reason I set this up is because now we have a number, and we can relate this back to the magnitude of the velocity at a. And once we can figure out this number, we can plug it back into our y-axis equations. Now we'll have our 3 out of 5 variables. That's why I have with this whole through this whole process of setting up through x. Alright. So, basically, what's going to happen here is we can divide by cosine of 37 on both sides. That cancels that out. We're going to have to divide by cosine 37 over here. Now if you go ahead and plug this into your calculators, what you're going to end up getting here is that, this is actually going to be 10. What's going to be it's a 10.01. We can we you can just round it to 10. So this is actually going to be the magnitude of the initial velocity, v_a. Now we can take this, and we can basically just plug it back into this over here. So this is going to be 10 times the sine of 37, and we know that 10 times the sine of 37 actually ends up being 6. So now we have here is we have 1, 2, 3 out of 5 variables, which means we can ignore the, the final velocity. And now we can go ahead and pick an equation to solve. A lot of steps here, and there's a lot of sort of going back and forth. But ultimately, we're just going back to whatever we do with, you know, what we would normally do with projectile motion problems. So we go over here, and the equation that ignores the final velocity is actually going to be equation number 3. So that's going to be the one that we're going to pick. Alright? So our final equation is going to be delta y from a to d is equal to the initial velocity. So this is going to be v_a_y, which we know, times the time delta a from t to d, plus 1 half, we have acceleration, and then times delta t from a to d, and that's going to be squared. Alright? So a lot of variables, a lot of little sort of little subscripts and letters. Really, these are all just numbers that we know. So delta a sorry, delta y from a to d is equal to the initial velocity. We know that's just 6. That's 6 times the delta t, which is just 3, so 6 times 3, plus 1 half. This is negative 9.8, and then this is just going to be 3 squared. When you plug this in, what you're going to get is you're going to get negative 26.1 meters. Alright? So is that our final answer? Well, no. Our final answer is not delta y from a to d. This is actually just oh, sorry. Our final answer is not negative 26.1. This is actually just basically how much the ball has fallen in flight. And this actually kind of makes sense. It's fallen because we were actually told that the ball is going to hit the building or the window at a lower height than it started from. So that delta y should be negative. Alright? So now to figure out our height above the ground, I'm actually going to go ahead and pick sort of different colors for this instead of doing this in green. This is actually just going to be I'll do this, like, in purple or something. Right? So that's going to be the height that it falls because our final answer is actually going to be the height of the window. So just plug this back in. It's going to be 50 plus our delta y from a to d, which is negative 26.1. And that actually works out to 23.9. But we're going to round it, and this is going to be 24 meters. And this is our final answer. Alright? So if we round our final answer, it's going to be 24 meters, and that's answer choice c. Hopefully, that made sense. Thanks for watching, and let me know if you have any questions.

Here’s what students ask on this topic:

What is non-symmetrical projectile motion?

Non-symmetrical projectile motion occurs when an object is launched from a height different from where it lands. This results in a trajectory that is not symmetrical. For example, if an object is launched from a cliff and lands on the ground, the motion is non-symmetrical. The key to solving these problems is to sketch the paths in the x and y axes, identify the target variables, and choose appropriate intervals for calculations. Understanding these principles helps simplify the equations and enhances comprehension of the motion dynamics.

How do you solve non-symmetrical upward launch problems?

To solve non-symmetrical upward launch problems, follow these steps: 1) Sketch the trajectory and draw the paths in the x and y axes. 2) Identify the target variable, such as the vertical component of velocity just before impact. 3) Choose the appropriate interval for calculations. For example, if calculating the vertical component of velocity, use the equation $v_d=v_a+a_{y}t$. This approach simplifies the problem and ensures accurate results.

What equations are used in non-symmetrical projectile motion?

In non-symmetrical projectile motion, the key equations used include: 1) $v_{\mathrm{dy}}=v_{\mathrm{ay}}+a_{y}t$ for vertical velocity, 2) $v_{\mathrm{dy}}2=v_{\mathrm{ay}}2+2a_y\mathrm{\Delta y}$ for vertical displacement, and 3) $\mathrm{\Delta y}=v_{\mathrm{ay}}t+0.5a_{y}t2$ for displacement over time. These equations help in calculating various parameters like velocity, displacement, and time.

How do you determine the vertical component of velocity just before impact in a non-symmetrical launch?

To determine the vertical component of velocity just before impact in a non-symmetrical launch, follow these steps: 1) Identify the initial vertical velocity $v_{\mathrm{ay}}$. 2) Use the equation $v_{\mathrm{dy}}2=v_{\mathrm{ay}}2+2a_y\mathrm{\Delta y}$ to find the final vertical velocity $v_{\mathrm{dy}}$. 3) Plug in the values for initial velocity, acceleration due to gravity, and vertical displacement. This will give you the vertical component of velocity just before the object hits the ground.

What is the importance of choosing the right interval in non-symmetrical projectile motion problems?

Choosing the right interval in non-symmetrical projectile motion problems is crucial for simplifying calculations and ensuring accuracy. The interval determines which variables and equations you will use. For example, using the interval from the maximum height to the ground can simplify calculations because the vertical velocity at the peak is zero. This reduces the number of variables and makes the equations easier to solve. Always try to choose an interval that includes known values and simplifies the problem.