Special Equations in Symmetrical Launches - Online Tutor, Practice Problems & Exam Prep

In symmetrical projectile motion, the total time of flight can be calculated using the equation tac = 2 × v × sin(θ) / g. The horizontal range is determined by R = v × cosθ⁢t or R = v × 2sin(2θ) / g. Maximum range occurs at a launch angle of 45 degrees, while complementary angles yield the same range. Understanding these principles is crucial for solving projectile motion problems effectively.

Using Special Equations in Symmetrical Launch Problems

Hey, guys. So in the last video, we were introduced to symmetrical launches, which is a type of projectile motion in which the final height equals the initial height. We saw some special properties of those kinds of motions. The fact that t_up = t_down and we saw that v_up = -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 your professor allows you to, and these are equations that are going to make symmetrical launches even easier to solve. So I want to show you how that works in this video. We're just going to get right to the example.

We've got a catapult that launches a projectile. We've got the launch speed and the angle. We're told that this v₀ here, which is just VA, is equal to 100, and we're told that the launch angle θ_a is 53 degrees. And we're going to find in this first part here the time that it takes for the projectile 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 2 symmetrical pieces. We found what 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 that we'll be covering, and it's basically just 2v₀sinθg. All you need to know is just the initial velocity and the launch angle, and you can actually figure out the time of flight for a symmetrical launch. This only 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, equations like v_up = -v_down. I want to show you this really quickly. In the A to C interval here if you use equation number 1 from your UAM equations, you have that the final velocity v_cy is v_ay plus a_y times t_ac. But what we know from symmetry is that if you have the initial launch velocity, if you have the y component, then when it gets to point C, it's going to be the same exact number except it's going to be downwards and so it's going to be negative. So v_ay, or sorry, v_cy is equal to negative v_ay. So if we go back to this problem here, we can actually just replace this v_cy with negative v_ay, right, because of this property over here. And this is v_ay plus a_y t_ac. So this actually just going to become 2v_ay-g. So that means that t_ac is just Well, what is v_ay? Well, if you have the launch velocity and you have the angle, then v_ay, v_ax is just v₀ cos θ and v_ay is just v₀ sin θ. So that means that our equation just becomes 2v₀sinθg, and this is actually just the equation that we just, we just arrived at over here. So what we can use is we can just use this equation straight from the get-go and say the t_ac is 2×100×sin(53°)9.8. That gives us the entire time of flight. We don't have to break this thing up. And so if you actually plug this in, you're going to get 16.3 seconds, and that's the answer. So that is the answer to the total time of flight. So let's move on now to the horizontal range of the projectile.

So what does that mean? Well, the horizontal range is, another 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 C is the whole time, then the horizontal range is just going to be R which is equal to Δx from A to C. It's going to be the entire distance that the horizontal that the projectile covers horizontally. So R which equals Δx from A to C. If we were to set up our equations, the only equation that we use in the x-axis is just v_x times t from A to C. So what 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 that straight into the equation. So in general, what happens is if you do have the total time of flight, if you have t_ac, then your range equation is just going to be Δx from A to C. It's just going to 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 R, which is still equal to Δx from A to C, just has a different form and it's going to be v₀2×sin2θg. So what I want to 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 the velocity is going to be v₀ times cos θ. So it's going to 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 t_ac, that's 16.3. So that's just going to be 980 meters. So that's what we get for the range. However, if we didn't already pick out the time of flight, and if we only just had the launch angle, the launch velocity, and the angle, then we could just plug it straight into the range equation. And so what we would do is we'd have to plug in the magnitude of the initial velocity, not the components. So this would be 100 squared times the sine of 2 θ. That's 2 times the launch angle. So we're going to do 2 times 53 degrees and then divided by 9.8. That's g. And if you plug this in, you're actually going to get 980 again. So we're going to get the same exact answer, and that's because really both of these equations mean the same thing. They're equal to each other. Alright? So that's how you use this range equation.

Here's just, you know, so you can memorize it or you use it on your homework and test questions and stuff like that. So there are a couple of things you should know about this range equation. The first is that it's maximum when the launch angle equals 45 degrees. So remember that this equation only depends on the initial. So what that means here is I've got this graph of these projectiles that are all thrown with the same exact v₀, the same velocity, and the one that goes the farthest is going to be the one where it's launched at 45 degrees. Anything less than or greater than 45 degrees is going to give you a smaller range, which actually brings me to my second point. Complementary angles, meaning θ₁ and θ₂ 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 are thrown with the same v₀s, then what happens is if I were to throw this projectile at 30 degrees rather than 45, then it would go it wouldn't go as high and it would also fall a little bit shorter than the maximum range at 45 degrees. If I were to instead launch 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're going to find the other launch angle that gives the same exact range as before, the 980 meters. So if we already have one of our angles and we know that θ₁ is equal to 53 degrees, and θ₂ is just going to be 90 minus θ₁. So it's going to be 90 minus 53 degrees, and that's going to be 37. The way you can test that is you can just plug R you can plug in θ equals 37 into your range equation, use the same exact numbers, so we're still going to use the same 100 squared times the sine of 2 times 37 now. If we divide this by 9.8, we're going to get 980 meters. So we're going to 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.

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°

A Long Jump on Planet X

Hey, guys. Let's check out this problem here involving a long jumper on some unknown planet. So we're told some information about their launch speed and the distance they can cover, and we're going to calculate the gravitational acceleration for this planet. So before I go ahead and draw and, you know, 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 going to jump upwards and then they're going to return back to the ground again which is great because we know we're dealing with a symmetrical launch, so we're going to be able to use our special equations to solve this problem. So we're told the initial launch speed at this \( v_0 \), which is 6. And let's just go ahead and draw our paths in the x and y. So this is my x. This is my y path here. This is a, b, and then back down to c 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're going to look at our 2 special equations here. We 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 \( v_0 \), theta, and then \( g \). Right? So we have \( v_0 \), theta, and \( g \). It's the same unknowns here. So, in order to figure out which one of these equations I'm going to use, I'm going to take a look at 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 that their maximum distance that they can cover turns out to be 9 meters. So this whole horizontal distance here, which is \( r \), is equal to 9 meters. This actually happens to be the maximum horizontal distance covered. So because I know some information about the range, I'm actually just going to use my range equation. So here's what I'm going to do. I have \( r_{\text{symmetry}} \). So \( r_{\text{symmetry}} \) is just equal to \( v_0^2 \times \sin(2\theta) / g \). So what I'm looking for here is I'm looking for the gravitational acceleration. Now I know what the range is, I know what our symmetry is. Basically, these are 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 then I'll be able to solve this equation because I only have one unknown. So what am I told here? How do I figure out theta? Well, the other thing I'm told about this range is that the maximum distance possible to cover is equal to 9 meters. 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 theta angle is equal to 45 degrees. So what 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 for \( g \). We're just going to use this equation here and then flip basically trade places between \( g \) and \( r \), which is going to divide \( g \) over to the other side and then, or multiply \( g \) over to the other side and then divide the \( r_{\text{symmetry}} \) down. So our equation just becomes \( g = (v_0^2 \times \sin(2 \times 45)) / r_{\text{symmetry}} \). So that means that our \( g \) ends up being, this is going to be \( 6^2 \times \sin(2 \times 45) / 9 \). And if you go ahead and work this out, what happens is this just turns into 1 and this is 36 over 9 which ends up being 4 meters per second squared. So that is our answer. 4 meters 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.

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

Here’s what students ask on this topic:

What is the total time of flight equation for symmetrical projectile motion?

The total time of flight for symmetrical projectile motion can be calculated using the equation:

$t$_ac=2v⁢sin⁢(θ)g

where $v$ is the initial velocity, $\theta$ is the launch angle, and $g$ is the acceleration due to gravity. This equation simplifies the process of finding the total time a projectile spends in the air when the initial and final heights are the same.

How do you calculate the horizontal range of a projectile in symmetrical motion?

The horizontal range $R$ of a projectile in symmetrical motion can be calculated using two methods:

1. Using the total time of flight:

$R=v\cos \left(\theta \right)t$_ac

2. Using the range equation:

$R=\frac{v^2\sin \left(2\theta \right)}{g}$

In both equations, $v$ is the initial velocity, $\theta$ is the launch angle, and $g$ is the acceleration due to gravity. Both methods will yield the same result for the horizontal range.

Why is the range of a projectile maximum at a 45-degree launch angle?

The range of a projectile is maximum at a 45-degree launch angle because this angle optimizes the balance between the horizontal and vertical components of the initial velocity. At 45 degrees, the sine and cosine functions are equal, maximizing the product $\sin \left(2\theta \right)$ in the range equation:

$R=\frac{v^2\sin \left(2\theta \right)}{g}$

Since $\sin (90°)$ is 1, the range is maximized. Any deviation from 45 degrees results in a smaller product, thus reducing the range.

What are complementary angles in projectile motion, and how do they affect the range?

Complementary angles in projectile motion are two angles that add up to 90 degrees, such as 30 degrees and 60 degrees. These angles yield the same horizontal range for a given initial velocity. This is because the sine of twice the angle is the same for both angles:

$\sin (230°)=\sin (260°)$

Thus, the range equation:

$R=\frac{v^2\sin \left(2\theta \right)}{g}$

yields the same result for both complementary angles, demonstrating that they produce identical ranges.

How do you derive the total time of flight equation for symmetrical projectile motion?

To derive the total time of flight equation for symmetrical projectile motion, start with the vertical motion equation:

$v$_cy=v_ay+a_y⁢t_ac

For symmetrical motion, the final vertical velocity $v$_cy is the negative of the initial vertical velocity $v$_ay:

$v$_cy=-v_ay

Substitute and solve for $t$_ac:

$-v$_ay=v_ay+-g⁢t_ac

$-2v$_ay=-g⁢t_ac

$t$_ac=2v_ayg

Since $v$_ay=v⁢sin⁢(θ), the equation becomes:

$t$_ac=2v⁢sin⁢(θ)g