1

concept

## Solving Non-Symmetrical Upward Launch Problems

7m

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Hey, guys. So in the last couple videos we saw how to solve problems. It was an object is launched upwards and then returns to the same height from which was launched. This was called symmetrical launches. But this won't always happened. You might see some problems which obviously launched and they returned two different heights. There's really just two options You could landed a higher height or a lower height here. 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 were really just gonna use the same steps and equations as before. Um, this just a couple of new things here. So what would this look like if you were landed at a higher heights? 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 wal or a building or something like that. And if you were launching at a lower height, so I like to call these launch from high problems. Then it would look like this. He would go up, reach its peak, and then it would come back down again and then hit the ground again. All right, So before we get into this and next bullet point, I actually just want to go ahead and start the problem, because, really, we're just gonna use the same exact system to solve them. So we've got this potato we're gonna fire from a potato launcher. We're on top of a cliff. It's 20 m. And so before we go ahead and draw anything on, start working their variables and equations, we're just gonna solve the first step. We're gonna draw the path C, X and y. So we've got this object here and it's gonna go up and down like this. So if we could only move along the x axis, it would look like that. And the Y axis. We'd go up to the peak, they would come back down again towards the bottom. So what are points of interest? Remember 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 that happens in between. So what happens is this object is gonna go come back around again. It's gonna pass the heights from which it was initially launched and then continue on further. So this is gonna be be we'll call the 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 gonna be just like a symmetrical launch problem. But 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, so again, we're gonna get back to the second the second little bullet point right here. Um, let's just get right to the problem. So we've got the paths and X and y. Now we're gonna determine the target variable. What are we actually looking for here? We're looking for the vertical component of the potatoes velocity just before it hits the ground. So that's gonna be at point D remember, there's two components of the velocity. There's gonna be the X components. I'll call this VD X, which is just the X because it never changes. And then we're looking for V. D. Y. So this is gonna be we're looking for here. Remember, these two components combined to produce a two dimensional vector v d. But that's not what we're looking for. We're just looking for V D. Y. So we're looking for that target variable here. Now the next big question is, what interval are we going to use or remember we have a bunch of different points of interest now. There's lots of different options as to the interval that we could choose. So there's a couple of options here, and depending on which one you take, you still get the right answer. But there's a couple of options that are easier and harder. One thing that you might be thinking of is, if you have the initial velocity, which is V A, which is just equal to 30 and what you could uses, you could use a symmetry argument you could say, Well, if VA is 30 then I know that V. C is also going to be 30 except just gonna be pointing in the opposite direction so I could use my components V C X, which is V X and then whatever V. C Y is is just gonna be the negative of V A y. So you could use this interval just from C to D, and you absolutely could solve it that way. But I'm actually gonna warn you that there's gonna be a little bit more work on. You're gonna have to go and solve a bunch of other things, the easiest choice for these kinds of problems and in general, what you should try to dio if you're ever landing at a lower height, is you try to choose intervals in which you're trying to include point B, which is the maximum height. You're always gonna want to try to include this. Equated this into your equations because one thing that we know is that the UAE velocity, when it's at the peak, is equal to zero, and we're going to see how this simplifies our equations and our variables. All right, So we're gonna be looking at the variable at the interval from B to D. So we're looking for the y axis, So I need all my wife variables. So I've got my A y, which is always negative 9.8 the initial velocity. Well, the initial velocity in this interval from B d. D is actually gonna be V b y. And you know that. Zero. This is actually why this interval is actually really easy to use. And you should always try to use it because we've already unlocked. We've already gotten two out of the five variables that we need. We just need one more, and it will be able to solve our equations. So the final velocity is gonna be the velocity at point D, which is exactly what we're looking for. And then we've got Delta y for me to be our sorry Delta y 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 Delta y 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 this vertical displacement over here. So which number is that? We'll see if we take a look at the problem. We're told that the potato is launched from a 20 m high cliff. So is that the Delta Y from ability? Well, no, because that's only gonna be this piece right here. This vertical distance is the 20 not this whole entire thing over here. However, we're also told from the problem is that the potato reaches its maximum height 49.4 m above the ground. So this is the number that this line represents 49.4 m. So this is gonna be our Delta y from B to D. However, because this points downward, it's gonna pick up a negative sign. So that's Delta Delta y. That's gonna be 49 point. Whoops. It's being 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 already have three out of five variables here. All right, so we're just gonna use the one that ignores time. That's my ignored variable. And that's gonna be equation number two. So this is the final velocity, which is V D. Y squared is gonna be the initial velocity V b Y squared. Plus, uh, a Y times t uh, I'm sorry. This is gonna be plus to a y Delta y from B to D. Okay, so we already know that this y velocity is gonna be zero B B y. And so v d Y squared is just gonna be two times negative. 9.8 times negative, 49.4. So if you go ahead and work this out, we're gonna get that V d. Y is equal to the square roots off, and this is gonna be 900. Um, this is gonna be 960 which is just gonna be 31.1. However, there's two answers here. We could use 31.1, and it's gonna be a positive 31.1, which means that we're getting a velocity that points up. Or you could get a negative 31.1. Because, remember, we're taking the square root of a number here. We're solving for velocity that would correspond to a velocity that points downwards. So which one of these answers is correct? Well, if you take a look at DT y ved white points downwards. So that means that this answer is not the correct one. And instead, we're gonna use V. D. Y is equal to negative 31.1 m per second. That's the 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.

2

Problem

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

3

example

## Height of a Roof

8m

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Hey, guys, let's check out this problem here. We've got a child that's throwing a ball up to a rooftop. We're gonna figure out how high is the roof. So let's go ahead. Draw diagrams. So I've got the ground level like this on then the rooftop like this. So we've got this ball that's being thrown at some angle like this, and it's gonna go up. And what else we told, we're told that the ball is gonna reach its maximum height directly above the edge of the roof. So when we go to draw diagrams, that's actually really important. We draw out the correct way. What this is saying here is that the maximum height of the trajectory is gonna be right above the edge of the roof, and then it's gonna fall back down. And so then it's gonna fall back down, and we know that this is gonna be 3 m away from the edge of the roof. So let's just go ahead and stick to the steps. We're gonna draw the X and Y pass and the points of interest and start filling out everything we know about the problem. So in the X axis we're going here and on 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 gonna look like this and in the why it's gonna look like this. All right, so that's our passing 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. All right, It's basically all we know here. Eso Let's go ahead and move on to the second step. We're gonna determine the target variable. What are we looking for It. We're looking for how high the roof is. So we look at our diagram here with that with that distance would represent is what actually represent this vertical displacement here. And if you look at it, that's basically from the ground, which where we shoot it, which we were, 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 see here. So that's actually gonna be our target Variable Delta Y A to C is gonna be the height of the roof, which actually brings us to the third step, the interval and which equations we're gonna use. Well, if we're looking for a to see, then one interval we can choose is the interval from A to C. But what happens is if you just go ahead and write all these variables, you're gonna end of figuring out that you can't actually solve this. So I'm just gonna warn you right there. If you try to do it, you're gonna get stuck. So we're gonna need a different kind of approach for solving Delta y from A to C. Now, one way you could do this. You could actually break up the motion into several parts If you wanted the Delta y from A to see, 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 gonna 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 be here. We're gonna be able to use two out of five 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 gonna use the interval from A to B instead. So I've got my A y, which is negative 9.8. We've got my V not y, which is V A y, which is Let's see, actually, let me go ahead and finish out the rest of variables. V Y is V B y. 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? All right, so let's go through each of these variables V A y. So if we have the magnitude which is 13 and the direction 67.4 so you can figure out the not X, which is V a X, which is just 13 times the co sign of 67.4. And if you do this, you're gonna get five. If you do V not why which is V A. Y? This is gonna be 13 times the sign 67.4 and you'll get 12. So we know this is 12 were here. What about V B y. The final velocity will remember. That's an important point in our project on motion problems, because we know that v. B y. Basically the velocity at the top here is equal to zero. Once you hit the top of your pique, your velocity is instantaneously zero. So it's momentarily zero for that little point there. So it's zero. So that means that we have 3 to 5 variables and we're looking for this Delta Y from A to B, which means I could just go ahead and pick the equation that ignores time. So that's what we're gonna do here. So that's equation number two, which says that the final velocity V B Y squared is V a Y squared, plus to a Y times Delta y from A to B right. We know this is equal to zero. That's why it's important and useful. We know this is 12 squared, plus two times negative, 9.8 times delta y from eight to be. If you go ahead and work this out, you're gonna get Delta Y from A to B is equal to 7.35 m. So this is only just part of the peace. 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 we're gonna dio is now that we're done with the A to be interval. We could just move onto the next one. So you can say in the B two C interval. I'm just gonna write out the same variables again. Right? So I've got negative 9.8. Now, my initial velocity from the B two c interval is actually gonna be V B Y. That's my initial velocity. We know that zero. What about B y? The final velocity That's gonna be V C Y on. 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 gonna be what we're plugging over here. And we should expect it to be negative and then t from B to C. All right, so now it looks like we have two out of five variables, but I'm gonna need either the final velocity or I'm gonna need time. So remember, whenever I get stuck in the y axis, I'm just always gonna go over to the X axis. So the easiest one to solve in the X axis is gonna be the time. So I'm just gonna go over here and I'm gonna look for time. So If I'm looking for TBC, then I'm still just gonna look at the BC Interval here. I'm just gonna write out my equation. Delta X BTC 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 dio because we just figured out that in the in the last part we know that's equal to five. What about Delta Extra BTC? Remember that it's 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 m from the edge off the roof, and we know that the distance from the edge to the roof over here is gonna 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, both of these variables, this is just three and then we have five times t v c, which means that T V C 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 three out of five variables 12 and three. So we're gonna go ahead and pick the one that ignores the final velocity. So we get Delta Y from B to C. I'm gonna use equation number three this time. So Delta Y from B to C is V B y. That's the initial velocity times T V C, which we know zero, because remember, this is just equal to zero plus one half times negative, 9.8 times 0.6 squared, right? So I'm just gonna go ahead and fill those in, and what you get is that Delta Y from B to C is equal to negative one point 76. So this is the vertical displacement, and it totally makes sense that we got a negative number because remember that the vertical displacement between these two points because it points downwards should be a negative number. So you get negative 1.76 here. And so when you add these two things together, you're gonna get 56 m and that is your final answers. That is the height off the building. So that's answer choice. See? Alright, guys. So we can see here. Whenever. Whenever you get stuck using one interval, you can always basically just break it up into a combination of intervals on. Go ahead and figure out what your target variable is using that method. All right, so let me know if you guys have any questions and I'll see you in the next one.

4

example

## Catapult Siege

5m

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Hey, guys, let's work out this problem together. So we got this cattle Paul that's launching a stone. It's gonna go up and then it's gonna hit a castle wall. We're trying to get the direction of the stones velocity, so let's just go ahead and stick to the steps. We're gonna 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's two different ways to draw the path off the stone. One possibility is that if you're gonna hit the castle wall while still on the way upwards, in which the final velocity is gonna be upwards like this or the other situation is that the cattle pump 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 gonna draw the past in the X and y. So the ex path is gonna look the same for both of them. It's just that one of them is gonna be a little bit shorter, but the white pass might look different. The white path of this one is just gonna go straight up and for this one is gonna go up and then come back down again. So when we label our points of interest is actually two possibilities, you could just be going from A to B, in which case your interval is gonna look like this A to B, or you could be going from a to B and then back down to see in which you hit the castle wall. So you're gonna 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 is launched to the point where it hits the wall. So I'm gonna call this interval even though I don't know which one a to see. So we're really looking for Here is we're looking for the angle that this velocity makes with the horizontal. And so remember those two components We've got V, C, X and V C Y. So you've got VC, VC, V, C, X and V C Y. And really, what happens is because we know the X velocity is never going to change. What's really gonna affect this is the why velocity. So let's go ahead and go to the second step here, which is our target variable. What are we looking for? Working for the direction which is data, and we're looking for the direction at point Cease. That's gonna be this variable over here. All right, So because data see is actually just a vector equation, it's just this tan universe over here. I'm gonna start with that. So we have tangent adverse off, and we need V C Y over V c X. So V C X is the easy one because remember that the X component of the velocity never changes. So what I can do here say that my V c X is just gonna be V ex throughout the whole entire motion. And that's actually just the initial velocity times the co sign of data. Right, which I know that's just 50 times the co sign of 56 I get 28. So that's never gonna change. It's always gonna be 28 throughout the entire motion there, V. C. Y is where it gets a little tricky because I don't know what this V c Y is, but I might have to go find it. I'm gonna have to pick an interval and then solve it using some, um, equations. So that's the third step. So we're interval. That we said we're gonna 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 wife is negative. 9.8. I've got V not why? Which is V A. Y which I can actually figure out here. My v a y is just gonna be, um or rather my v initial. Why is gonna be va y? And that's va V, not times the sine theta. So that's just 50 times the sign of 56. And that is 41 points five man. So 41.5. Okay, so we know this is 41.5, and then RV final Y is gonna be V C y. That's actually what we're trying to look for. And then see, we've got Delta y from A to C and then t from A to C. Now we do know that it takes six seconds for the stone to hit the castle walls. That's t A. C. And so that means that we have 3 to 5 variables and we picked the equation that ignores my delta. Why? So this equation here is actually gonna be equation number one, which says that the final velocity V. C. Y. Is v a y plus a Y times t a c so v c y equals our initial velocity in the Y axis 41. plus negative 9.8 times how long it was in the air. Six seconds. If you go ahead and work this out, you're gonna get negative 17 3 meters per second. So this is our why components of our velocity. Remember, this is not a 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 gonna do here is Theta C equals the tangent in verse. Off, we've got negative. 17.3 divided by 28. That's the X component. And if you go ahead and work this out, what you're gonna get is you're gonna get negative 31.7 degrees. So which one of our two situations that end up being well, because the y component end up being negative? It's actually going to be this situation over here. This was the correct one. And so this angle over here Thatta see, because it's negative means it's being below the horizontal. So your fantasy is negative 31.7 degrees, and that is your final answer. All right, so that means that brings us to answer choice A. And let me know if you guys have any questions about that one

5

concept

## Using Single Intervals in Positive Launch Problems

7m

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Hey, guys. So by now we've gone through all the steps to solve project on motion problems for any given problem, we're just going to sketch the trajectory and draw its past in the X and Y. And then once we figure out our target variable, we're gonna pick our intervals and then start using our equations. And what we've seen in Project on Motion is that oftentimes 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 talk to calculate the total amount of time from initial to final. We'll actually 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 added to the time that it takes to go from B to D. So just adding those two smaller times there. But another option is like I also just try to calculate the time from a to see because I know that is symmetrical, and I can try to use some of the symmetry points there. So that's Delta A from Delta T from A to C and then attitude 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 oftentimes, the easiest approach to solving these problems is just by using a single interval, going from a all the way down to D, you could absolutely do that. And in fact, that's gonna 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 gonna 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 in the angle. We're gonna 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 gonna draw the past in the X and y. So we're gonna go ahead and have passed 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 it's B, C and D. And so now that's the first step. The second step is we're gonna figure out the target variable. So in this, uh, first part here, we're gonna calculate what the vertical components of the velocity at the point where it hits the ground. That's point D. So remember, at any point there's two components of the velocity. There's VD X. There's the X component and the white components. That's V. D. Y. And both of these combined to form a two dimensional vector using the Pythagorean theorem V. D. So which one of these variables we're looking for? We're looking for the vertical components that is V. D. Y. So that's our target variable. So now just comes to what equation or what interval are we gonna use? Well, remember that these problems were gonna try as best as we can to use the interval from a D. D. If we have enough information, if we get stuck and we turns out we don't. We can always just come back and then picked different intervals. So let's try to use the interval from A to D. All right, so I'm looking for a y axis variable. So that means I'm just gonna use I'm gonna list out all of my wife variables. Right. So this is gonna be a y, which is negative. 9.8. And that's regardless of the interval that we're using. Remember a Y. It was always negative Jeans negative 9.8. No matter which interval you're gonna choose. So our initial velocity is our velocity at point A. Our final velocity is V D y. That's actually we're looking for, You know, 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 v a y. And that is I configure that as long as I know the initial velocity, which is V A, which is 100 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 V a X, which is just Vieques throughout the entire thing the entire motion will never change is times the co sign of 30 and that's 86.6. That number is never gonna change throughout the whole motion with y Velocity, on the other hand, will change. That's V A y, and that will be 100 times the sign of 30. And that's gonna be 50. So that's our variable. There we have V A Y 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 d. D. Well, the only other thing that we're told in the problem is that this cannon is fired from a m cliff so that 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 too deep. 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 gonna be 40 downwards again, forget about what happens in between. It doesn't matter. That goes up and then comes back down again. It's gonna be negative. 40. So that's negative. 40 over here. And so therefore, we have our three out of five variables. I've got 12 and three, and I'm gonna pick the equation that ignores time. So that's equation number two. So I've got V d Y squared equals V A Y squared plus to a Y times delta y from A to B I have all those numbers and I'm just go ahead and plug so my V d is gonna be the square roots of. And then I have 50 squared plus two times negative, 9.8 times negative, 40. If you go ahead and work this out, what you're gonna get is you're gonna 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 nine, there's two answers positive and negative. Three. Both of those things, when you square them will both equal nine. So which one of these makes sense? Well, a positive 57.3. Meaning it would mean that the velocity is going up and a negative 57.3 meetings means that it would go down. So obviously at point d r velocity component is gonna be downwards. So therefore, this is not the right answer. And this negative 57.3 degrees is our right answer. All right, let's let's move on. So this point, part B Now, now we'll be 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 were still just gonna try to use a two D. In fact, T a d was the was the variable that we initially ignored in this first part. But now that we know V. D. Y, now we know this is just negative 57.3. Now we actually have four out of the five variables and so we can actually use any one of the equations to solve 40. Any one of these equations will be, uh is totally fine for solving the time. So what we're gonna do is we're gonna use equation number one because it has the least amount of terms. It's the simplest equation to use. So this just says that v. D. Y here in our same interval at a D is just gonna be v a y Plus on this is gonna be negative. Whoops. Plus a Y times t eight a d. So we've got negative. 57.3 equals 50 plus negative 9.8 times t a to d we go ahead and move this over, we're gonna get negative. One of 7.3 equals negative, 9.8 times ta two d. And so we're gonna get negative 107.3, divided by negative 9.8. The negatives will cancel, and then you'll just get tea 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 gonna make your whole life a lot easier if you can try to do this is best a zoo much as you can. Anyway. That's it for this one. Guys, let me know if you have any questions.

6

Problem

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

Additional resources for Positive (Upward) Launch

PRACTICE PROBLEMS AND ACTIVITIES (14)

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- In a carnival booth, you can win a stuffed giraffe if you toss a quarter into a small dish. The dish is on a s...
- In a carnival booth, you can win a stuffed giraffe if you toss a quarter into a small dish. The dish is on a s...
- A shot putter releases the shot some distance above the level ground with a velocity of 12.0 m/s, 51.0° above ...
- A shot putter releases the shot some distance above the level ground with a velocity of 12.0 m/s, 51.0° above ...