Parametric equations for projectile motion
About this video
A water balloon is launched from 1 m high at 25 m/s, 35 degrees above the ground, toward a 2 m tall target 55 m away. The velocity is split into components 25cos(35°) and 25sin(35°), giving x(t) = 25cos(35°)t and y(t) = 25sin(35°)t - 4.9t^2 + 1. Solving x(t) = 55 gives t ≈ 2.686 s, and then y ≈ 4.2 m, so the balloon clears the target.
You'll be able to
- split a launch velocity into horizontal and vertical components using cosine and sine.
- write parametric equations x(t) and y(t) for projectile motion, including gravity and a starting height.
- find the time when x reaches a given distance and use it to find the height y.
Key words
- vector
- A quantity with both a size (magnitude) and a direction, like a velocity of 25 m/s at 35 degrees.
- component
- The horizontal or vertical part of a vector.
- parametric equations
- A pair of equations giving x and y each as functions of time t.
- acceleration due to gravity
- About 9.8 m/s^2 downward; it slows the balloon going up and speeds it coming down.
- SOH CAH TOA
- Memory aid: sine = opposite/hypotenuse, cosine = adjacent/hypotenuse, tangent = opposite/adjacent.
Transcript
- [Instructor] We're told David uses a catapult to launch a water balloon from a height of one meter with an initial velocity of 25 meters per second at an angle of 35 degrees from the ground. He's trying to launch the water balloon over a target that is 55 meters away and is two meters tall, right?
Two parametric equations that model the path of the water balloon and what is the height of the water balloon when it reaches the target? So pause this video, have a go at that, and then we'll do it together. All right, so let's work through this together. And this is gonna involve a little bit of physics.
If you wanna go deeper on the physics or prove some of the formulas, I encourage you to look at physics on Khan Academy. But we have an initial velocity of 25 meters per second at an angle of 35 degrees. Some of you might be already familiar with the concept of a vector, but that's one way of representing something that has a magnitude and a direction.
So in this case, our magnitude is 25. We're gonna assume everything is in meters per second when we're talking about velocity. And it's direction, it's at an angle of 35 degrees with the horizontal. Now what we want to do to set up our parametric equations is break down this velocity in terms of its horizontal or its X component and it's vertical component.
And then we can set up its x position as a function of time and it's y position as a function of time. Those will be our two parametric equations. So let's think about the velocity and the x direction. You could think of breaking up that vector, that arrow right over there into two components.
That would be the X component. And then maybe in, let me pick another color here that I haven't used yet. Maybe in purple I will have the, oh, that's not purple, I'll go with it anyway, that is the Y component. Now, we know from trigonometry, this is a right triangle and we can just use, actually not even the unit circle, we could even go all the way back to our SOHCAHTOA definitions of sin, cosin, et cetera.
This 25 meters per second, this 25 is the length, the magnitude of that hypotenuse. So what we have right over here, the horizontal component, that will be the hypotenuse times the cosine of this angle, cosin of 35 degrees. We know that cosine is adjacent over hypotenuse.
And then from that, you can get this right over here. And then similarly, the vertical component is going to be 25 times the sin of 35 degrees. Now, just from this, we're very close to setting up what our x position is as a function of time. We can say that x as a function of time is just going to be our velocity in the x direction.
So 25 times cosine of 35 degrees, whatever this number is, we could evaluate it with a calculator times time. We're assuming we don't have any acceleration, we're we're assuming there's no friction or air resistance. Maybe we're doing this on a planet someplace or on a moon that has no atmosphere.
But this is oftentimes an assumption we make when we're doing these introductory types physics problems. So this is our first parametric equation. Now let's think about it in the y direction. Now this is going to get a little bit more involved. And I wanna get that same color.
Yeah, there you go. Is that the same color? No, not quite. As you can tell, I'm sometimes I'm impaired by these colors a little bit. All right. So y of t, yeah, that's the same color is going to be equal to, if we didn't have any gravity and any air resistance, we would just take the why component of velocity and multiply it by time.
But we know we have a gravity. And I'm not gonna prove the equations here, we do this in other videos on Khan Academy, but your y as a function of time is going to be your initial velocity, 25 times sin of 35 degrees times time. But we're going to be taking away from that initial velocity until we get to zero.
If we just imagine what we're going to be doing, our trajectory for this, what are we doing? A water balloon. It's going to go up and then it's going to go down. Or if you just imagined it's gonna take a parabolic trajectory like that. And if you just imagined it's y velocity, it'll start really high going upward and then it's going to, so it's gonna start high, then it's gonna slow down, slow down, slow down, slow down, and then it's going to start increasing.
But then as it goes down in the other direction. And so we have to incorporate gravity here. And the way that we incorporate gravity here in the equations is we take the acceleration of gravity, which is 9.8 meters per second squared over two times t squared. Now if we were just starting at y of zero zero, then this would be it.
But we know our at y of zero when t is zero, we're starting at a height of one meter. So we wanna make sure that that's there too. With x of t, I don't have a term like that because we're just gonna assume that x of zero is zero and then we're gonna see how long it took us to move 55 meters or to make x equal to 55.
But there you have it. I at least have my two parametric equations. Now the next step is to figure out what is the height of the ball when it reaches the target. So what I need to do is figure out at what time is x of t or is X equal to 55? And then I could put that time back into this equation, the second parametric equation to figure out my height then.
So let's do that. So I'll go back to yellow. So 55 meters, I'm assuming everything's in meters and meters per second is equal to 25 times the cosine of 35 degrees times t. Now if we wanna solve for t, we would get t is equal to 55 over 25 cosine of 35 degrees.
I guess we could simplify this a little bit. This is the same thing as 11 over five. But let's figure out what that is or at least what that is approximately going to be. Get a calculator out. Alright, so let's first make sure that we are in degree mode. If I click rad here, we go to the radian mode.
So let's go back into degree mode here. And so let's go to 35 degrees. I'm taking the cosine of it. Now let me multiply it by five. I'm working on the denominator right over here. And then let me take the reciprocal of that or one over that and then multiply that times 11.
Times 11. And so that's approximately if I were to round, well, I'm gonna round, eventual answer I want round to the nearest 10th. But it doesn't hurt to have a little bit more precision in the in-between answers. So let's just call that 2.686. So that's approximately 2.686 seconds.
Now what we wanna do now is take this value and evaluate y at that time. So what we want to do, we want to figure out y of, I'm running out of space here, y of, I'm gonna write small, of 2.686 is going to be equal to 25 sin of 35 degrees. All of that times 2.686, and then minus, this is the same thing as 4.9 times 2.686 squared and then plus one.
So let me just get my calculator out again. 35 sin, which is that times 25. Times 25. This is essentially our initial velocity in the y direction. If we wanted to write it out here, it'd be roughly 14.3 or 14.34 meters per second going straight up in the y direction.
But we're gonna multiply that, we're gonna see what our height is at 2.86 seconds. So times 2.686 is equal to that. So it's 38.5, at least this first term. And then from that, I want to subtract and let me open that parentheses and I could get 2.686 squared times 4.9.
Times 4.9. And if I were to close, I got the 4.9, that's just 9.8 divided by two. If I close the parentheses, I get that. And so this is going to be equal to 3.16 and then I have to add that one at the end. So plus one is 4.164. If I'm rounding to the nearest 10th, it's approximately 4.2 meters.
So all of this business is approximately 4.2 meters. So it's clear that we are actually clearing our target.
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