Special Equations in Symmetrical Launches - Video Tutorials & Practice Problems

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concept

Using Special Equations in Symmetrical Launch Problems

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7m

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Hey, guys. So in the last year, we were introduced to symmetrical launches, which is a type of projectile motion in which the final heights equals the initial heights. We saw some special properties of those kinds of motions. The fact that tee up equals t down. And we saw that V up equals the negative of V down various things that come from symmetry. But on top of these special properties that apply to symmetrical launches, there are also some special equations that you may be allowed to use if you're Professor allows you to and these air equations, they're gonna make symmetrical launches even easier to solve. So I want to show you how that works in this video. We're just gonna get right to the example. So we've got a catapult that launches a projectile. We've got the launch speed and the angle. We're told that this V not here, which is just V A, is equal to 100. We're told that the launch angle Fate A is 53 degrees and we're gonna find in this first part here the time that it takes for the project out to hit the ground again. So what that means is we have to figure out the time that it takes for the projectile to go from point A to point C. Now, the way that we did this before in the last video was that we basically just broke this up into two symmetrical pieces we found with the time up was and then we saw that the time up equals the time down and so we were able to just double it. However, a more straightforward approach is actually just called the total time of flight equation. This is one of the special equations it'll be covering, and it's basically just to v not signed data over G. All you need to know is just the initial velocity and launch angle, and you can actually figure out the time of flights for a symmetrical launch. This Onley works for symmetrical launch equations and problems. So just make sure you understand that this equation actually is a consequence of one of our symmetry. Um, equations like V up equals negative V down. I want to show you this really quickly in the A to see interval here. If you use equation number one from your, um equations you have that? The final velocity V c. Y is v a y plus a white times t a c. But what we know from symmetry is that if you have the initial launch velocity, if you have the white component, then when it gets to point C, it's gonna be the same exact number. Except it's gonna be downwards. And so it's gonna be negative. So v a y sorry V C Y is equal to negative v A y. So if you go back to this problem here, we can actually just replace this V c y with negative v A y right because of this property over here and this va y plus a Y t a c So this actually just gonna become to V A Y were divided by negative G because that's what the acceleration of the Y component is. And so these negatives were gonna cancel, and this is gonna be ta see? So that means that T A C is just what is V A y? Well, if you have the launch velocity and you have the angle than va Y via X is just V, not cosine theta and va y is just the not signed data. So that means that our equation just becomes to V, not sine theta over G. And this is actually just the equation that we just we just arrived at over here, so we can use is we could just use this equation straight from the get go and say the t A c is to be not sine theta over G. And so therefore, it's two times 100 times the sign of 53 degrees divided by 9.8. That gives us the entire time of flight. We don't have to break this thing up. And so if you actually plugged this and you're gonna get 16.3 seconds and that's the answer so that is the answer to the total time flights. So let's move on now to the horizontal range of the projectile. So what does that mean? Well, the horizontal range is under the word for that is actually just the total horizontal displacement. It's how much distance the projectile covers from start to finish. So, just like this whole entire interval here from a to seize the whole time, then the horizontal range, it's just gonna be our which is equal to Delta X from eight to see, it's gonna be the entire, uh, distance that the horizontal that the projectile covers horizontally. So our which equals Delta X from A to see if you were set up our equations, The only equation that we use in the X axis is just v x times t from a to see So we can use is now that we've actually figured out this time the 16.3 seconds we can actually just go ahead and plug it straight into the equation. So in general, what happens is if you do have the total time of flight if you have t A C and your range equation is just gonna be Delta X for me to see, it's just gonna be velocity times time. It's basically just the normal, you know, equation that we've been seeing so far. However, if you didn't already calculate that total time of flights, then there's actually another equation we can use. It's called the Range Equation, and so our which is still equals, and Delta Extra Odyssey just has a different form, and it's gonna be V not square times. The sign of tooth data over G. So what I wanna do in this problem is I want to show you that using both of these equations will still get the same exact number. So, for instance, what is the X component of the velocity? Well, we'll go back here and say that the X component of velocity is gonna be V not times cosine, theta. So it's gonna be 100 times the cosine of 53. And if you go ahead and do that, what you should get is you should get 60. So if we plug it into our problem here, we've got 60 times and then we've got the TA see. That's 16.3. So that's just gonna be 980 m. So that's what we get for the range. However, if we didn't already think at the time of flights, and if we Onley just have the launch angle, the launch velocity and the angle that we could just plug it straight into the range equation. And so what we would do is we have to plug in the magnitude of the initial velocity, not the components, So this would be 100 squared times the sign of two theta That's two times the launch angle. So we're gonna do two times 53 degrees and then divided by 9.8. That's G. And if you plug this in, you're actually gonna get 980 again. So we're gonna get the same exact answer. And that's because really, both of these equations mean the same thing. They're equal to each other. All right, so that's why you use this range equation, um, is here is just, you know, so you could memorize it or you use it on your homeworks and test questions and stuff like that. So there's a couple things you should know about this range equation. The first is that its maximum when the launch angle equals 45 degrees. So remember that this equation Onley depends on the initial velocity and launch angle. So for any given value of V knots, you'll attain the maximum amount of range when the launch angles equal to degrees. So what that means here is I've got this graph of these projectiles that are all thrown with same exact V, not the same velocity, and the one that goes the farthest is gonna be the one where it's launched at 45 degrees. Anything less than or greater than 45 degrees is gonna give you a smaller range, which actually brings me to my second point complementary angles, meaning fate a one and faded to equal to 90 degrees, such as 30 and 60 Right, because 30 plus 60 equals 90 complementary angles will give you the same exact range for the same exact initial velocity. So again, if all these projectiles were thrown with the same V knots and what happens is, if I were to throw this projectile at 30 degrees rather than 45 then it would go. It wouldn't go is high, and it would also fall a little bit shorter than the maximum range at 45 degrees. If I would instead launched this at 60 degrees, it would go a little bit higher, but it would also fall to the same exact range as the 30 degree angle over here. So that brings us to the last part where you gonna find the other launch angle that gives the same exact range as before the 980 m. So if we already have one of our angles that we know that Data one is equal to degrees in Theta two is just gonna be 90 minus data one. So it's gonna be 90 minus 53 degrees. And that's gonna be 37 on the way. You could test. That is, you can just plug are you can plug in theta equals 37 into your range equation. Use the same exact numbers. So we're still gonna use the same 100 squared times the sign of two times 37. Now, if we divide this by 9.8, we're gonna get 980 m. So we're gonna get the exact same range, even though the launch angle is different. And that's because they're complementary angles. That's it for this one. Guys, let me know if you have any questions.

2

Problem

Problem

A frog leaves the ground with a speed of 15 m/s and stays in the air for 2.0 seconds. At what angle did the frog jump?

A

40.8°

B

9.4°

C

19.1°

3

example

A Long Jump on Planet X

Video duration:

3m

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Hey, guys, let's check out this problem here involving a long jumper on some unknown planet. So we told some information about their launch speed and the distance they can cover, and we're gonna calculate the gravitational acceleration for this planet. So before I go ahead and draw, you know, and work with any equations, let's just go ahead and draw a quick little sketch of what's going on here. So we have a long jumper who's gonna jump upwards, and then they're gonna return back to the ground again, which is great, because we know we're dealing with the symmetrical launch, so we're gonna be able to use our special equations to solve this problem. So we're told the initial launch speed at this Venus, which is six. And let's just go ahead, draw our paths in the X and y. So this is my ex. This is my Why path. Here, this is a B. And then back down to see again. This is point B. The maximum height. Okay, so what are we actually looking for here? We're looking for the gravitational acceleration on this planet, so that variable is G. So we look at our two special equations here. We've got one for the symmetrical time. And we also have one for the total horizontal displacement or the range. Now what happens is unfortunately, g pops up inside of both of these equations. So remember that both of these equations time and range depend on Venus feta and then G right, so we have enough data and G, it's the same unknowns here. So in order to figure out which one of these equations I'm gonna use, I'm gonna take a look. At what? What? What I know the most information about. So, for example, here we're not told anything about the amount of time that this long jumper spends in the air. However, we do know, we do know that their maximum distance that they can cover turns out to be 9 m. So this whole horizontal distance here, which is our is equal to 9 m. This actually happens to be the maximum horizontal distance covered. So because I know some information about the range, I'm actually just gonna use my range equation. So here's what I'm gonna dio I have are symmetry. So our symmetry is just equal to V, not squared times the sign of two theta divided by G. So I'm looking for here is in looking for the gravitational acceleration. Now I know what the range is. I know what our symmetry is. Basically, this is just the same numbers and I know what the initial launch speed is. All I need to do is I just need to figure out the launch angle and they will be able to solve this equation because I only have one unknown. So what do I What am I told here? How do I figure out data? Well, the other thing I'm told about this range is that the maximum distance possible to cover is equal to 9 m. And so remember that there's a special property about symmetrical launches. The maximum distance that you can cover for any given launch Velocity happens when the data angle is equal to 45 degrees. So that this problem is actually telling us or we can gather from this is that the launch angle is actually 45 degrees. Therefore, we do know what this launch angle is, and so therefore we can solve Fergie. We're just gonna use this equation here and then flip basically trade places between G. And are we just gonna divide you over to the other side and then or multiplied you over to the other side and then divide the are symmetry down. So our equation just becomes G equals V not square times the sign of tooth data two times data divided by our symmetry. So that means that RG ends up being This is gonna be six square times the sign of two times 45 divided by nine. And if you go ahead and work this out, what happens is this just turns into one and this is 36/9, which ends up being 4 m per second squared. So that is our answer. 4 m per second squared is the gravitational acceleration. So therefore we're looking at answer choice B. That's it for this one. Guys, let's keep going

4

Problem

Problem

A golf ball is hit at ground level at an angle of 31.9° above the horizontal. Its range is 257 m over a level green. What was the magnitude of the golf ball's initial velocity?

A

2807 m/s

B

69 m/s

C

95 m/s

D

53 m/s

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