Gerak Parabola • Part 1: Konsep, Skema, dan Rumus Gerak Parabola

Welcome to the channel of Jendela Science, the channel for those of you who want to understand the lessons of Mathematics, Physics and Chemistry in high school. In this video we will discuss the first part of the parabola movement, namely the scheme and formula of the parabola movement. Keep watching this video until the end. This is the scheme of the parabola movement, so if there is an object,

At first, it is placed on the ground, for example at point A. Then this object is thrown up with a speed of the initial V0 and an angle of alpha towards the horizontal. Or if you know it, it's called an elevation angle.

then, according to the Ruh's vector, this V0 can be divided into V0x and V0y. Remember, V0x is the one next to the alpha angle, which means V0 cos alpha. And V0y is the one across or in front of the alpha angle, which means V0y is equal to V0 sin alpha. If the object moves like this, then the object's path will be a parabola like this.

So, the example of a parabolic movement in everyday life is, for example, a bullet is fired from a gun, and it hits the target, then the bullet will pass like this. Or, for example, a football player kicks the ball, the ball is kicked, it goes up, then it goes down again. That is an example of a parabolic movement in everyday life. Here, in a parabolic movement, there are four important points that we will discuss one by one. The first point is point A, which is when the object is thrown.

The second point is point B. Point B is located at a certain angle, at a certain position. At point B, the speed of the object is v. And we can also divide v into vx and vy. vx is the horizontal component of the speed of the object, and vy is the vertical component of the speed of the object.

The third one is C point. C point is located at the top. If the shape is a parabola, it means there is a top. At this top, the magnitudes that we will calculate are Xmax and Ymax. Xmax is the horizontal distance from A to C, or the horizontal distance from the object starts to be thrown until the object is at the maximum point, the highest point.

while Ymax is the vertical distance or what we know as the maximum height of the object.

That's Ymax. Then the last point we need to pay attention to is D, where this object returns to the ground. The path of the object is like this, it starts moving, tilts upwards, then it tilts upwards until it reaches the highest point, then tilts downwards, and returns to the ground. This is D. The horizontal distance between A and D is called the farthest X, or the farthest distance reached by the object.

Actually, we can call this parabolic motion as a combination of two movements, namely horizontal and vertical movement. The principle of horizontal movement, or movement according to the x-axis, is GLB.

while the vertical movement or movement according to the sumu-ye is the vertical movement above or GVA you have already read in the last straight movement about vertical movement there is a vertical movement above that is when the object is thrown upwards with a certain speed then the object will rise then fall again like this well if the parabolic movement is like this if GVA, the object from here is thrown upwards then down again

Parabola movement is GVA while GLB, which means it is GVA while moving to the right, which means it goes up and down. Do you understand what I mean? So here, someone says that this parabola movement is a combination of two movements or a combination of two movements. Next, we discuss these four points one by one starting from point A.

In the initial position, or point A, the speed of the object, as I explained earlier, is V0. Then we divide V0 by the x-axis and the y-axis, into the horizontal and vertical components. The horizontal component of the initial speed, or V0x, is called V0 cos alpha, because V0x is at the side of the angle. While the vertical component of the initial speed, or V0y, is the same as V0 sin alpha, because it is at the side of the angle.

Then the position coordinates, so it's like this.

the base or the ground is the source X and this is the source Y if this is the source X and this is the source Y then the point A is located at 0.0, right? the coordinate is 0.0 except if the parabola movement does not start from the ground if it doesn't start from the ground, for example, the bullet was fired by a cannon, the cannon is not placed on the ground the cannon is placed on top of a tower or at a certain height so the Y0 is as high as the certain height, 2 meters or

4 meters or something like that. Or for example, the ball is not kicked from the ground, the ball is kicked from above the building. That means Y0 is the height of the building. Do you understand? This is the speed and the position coordinates. Okay, let's go to point B now. At point B, which is a certain point, this object has a speed of V that we can classify as a horizontal and vertical component, namely Vx and Vy.

Vx = V0x, because remember the principle, horizontally, the movement of the x-axis is glb, if glb means at any point the horizontal speed is the same, so Vx = V0x, V0x is what? V0 cos alpha, so Vx = V0 cos alpha,

What about the speed for the value of y? The speed for the value of y is according to GVA. What is the formula for GVA? The formula for GVA is Vt = V0 - Gt. We give the index y here, which means Vy = V0y - Gt. V0y is V0 sin alpha, which means the formula is Vy = V0 sin alpha - Gt.

Okay, then V is the resultant of Vx and Vy, so the formula is like you learn it in vector through the analytical method, right? At that time, the resultant or sigma f is equal to the root of sigma fx squared plus sigma fy squared. If here, V is equal to the root of Vx squared plus Vy squared. Then the position coordinates, we use the distance formula. If GLB, what is the distance formula?

S = V x T. If we replace S with X, X is the horizontal distance. Y is the vertical distance. Horizontal distance or X is equal to V x T. It means Vx x T. Since Vx is equal to V0x and V0x is V0 cos alpha, we can use X = Vx x T or X = V0 cos alpha x T. Or the horizontal component of speed multiplied by time.

Then what about Y? What is the formula for Y in GVA? Y = Y0 + V0T - 1/2GT^2 Now, V0 is not V0, but V0Y because it's only the vertical component. So Y = Y0 + V0YT - 1/2GT^2 V0Y is V0sinα, so Y = Y0 + V0sinαT - 1/2GT^2 So basically, this formula is derived from the GLB and GVA formulas.

There is nothing new that we have developed here. What you need to remember is that V horizontally is always equal to V0x, always V0 cos alpha. If V vertically, it changes, but V0y is always V0 sin alpha. Okay? Let's continue to the third point.

The third point is when the object is at the peak or at the maximum point. If the object is at the highest point or maximum point, then the speed of the object vertically or vertically is equal to zero. Because it's like a GVA at the maximum point. At the maximum point, the object is going up and down. It means that Vy is equal to zero.

It means that the current speed of the object is purely the horizontal speed. So V = Vx or V0x because it's GLB, equals V0 cos alpha. Okay? Usually, when you're asked to find the time it takes to reach the maximum point, that's T max. We put it in the formula Vy = 0, and T is T max. Let's try to make the formula. So Vy = 0.

Remember the formula for V_y? V_0y - G_t = 0. V_0y is V_0 sin alpha. V_0 sin alpha, main G_t, we move to the right side, so it's equal to G_t. It means T = V_0 sin alpha per G. T is Tmax, so we get this formula.

The first formula. Tmax = V0 sin alpha/g. So this is like an instant formula. If you use a principle, use Vy = 0. Same thing. After that, usually we are asked about the position coordinates. The position coordinates are Xmax or Ymax. Xmax is the horizontal distance from the starting point to the highest point. While Ymax is the vertical distance. This is more often asked about Ymax. Ymax is the vertical distance, which means the maximum height of the object.

So we put Tmax into the formula X and Y. What is the formula X? When it is at a certain point, X is equal to Vx times T. Vx is equal to V0x, which is equal to V0 cos alpha. Times T, T is Tmax, right? T is Tmax, which means V0 sin alpha per G. Okay? It is equal to V0 times V0, V0 squared. Sin alpha cos alpha per G.

So this is the formula from Xmax. Then, why does sin 2 alpha suddenly appear here? Because you have only learned the formula for sin 2 alpha in class 11. So I will give you the formula here first. So there is a trigonometric formula like this. sin 2 alpha equals 2 sin alpha cos alpha. So if it's only sin alpha cos alpha, then we move the two to the left. So it's half sin 2 alpha equals sin alpha cos alpha.

This formula can be modified to be X = Vn² sin α cos α = 1/2 sin 2 α, so sin 2 α and 1/2 is put below, so it becomes 2G. Next, we will prove the Ymax formula. The way to do it is to put Tmax into the formula Y.

Remember the formula for y at a certain point? y = y0 + v0y t - 1/2 g t^2. We leave the y0, plus v0y. v0y here is v0 sin alpha. So v0 sin alpha t - 1/2 g t^2. We leave the y0. Then v0 sin alpha remains

we replace T with Tmax, which is V0 sin alpha per G.

minus half G. Now, T squared here means T max squared. So, we square V0 sin alpha per G. Let's do it right away. V0 sin alpha per G in the square means V0 squared sin squared alpha per G squared. All of them are squared. So, it's the same as Y0 plus V0 sin alpha multiplied by V0 sin alpha, which means V0 squared sin squared alpha per G.

Minus, here G and G^2 can be subtracted, so this is subtracted with this one, so the one above is V0^2 sin^2α per the lower one, which means 2 times G, 2G. We equate the denominator, so the easiest thing is that we multiply V0^2 sin^2α by 2, the upper and lower. 2, this is 2, like this. So it's the same as Y0

Plus, here the names are the same for 2g. So we just need to subtract the upper one. 2V0^2 sin^2α - V0^2 sin^2α = what? It means V0^2 sin^2α, right?

per 2g. So we get the formula Ymax is Y0 plus V0 squared sine squared alpha per 2g. Okay, here Y0 is the same as in GVA, it is the initial height. The object is thrown at what height? If it's from the ground, it means Y0 is 0, just ignore it. Okay, understand? So we can prove this special formula or this instant formula through the principle here. Okay, let's continue to the last point, point D.

Point D is when this object reaches the ground. If it reaches the ground, then the formula is like this. v_x remains v_0x, v_y remains v_0y-g-t, and the result is the same, root of v_x^2 + v_y^2. Then the coordinate of the position, y must be 0, because it will definitely reach the ground. Well, t is called t_ground, the time it takes to reach the ground.

Especially if the motion of the parabola starts from the ground and goes back to the ground, then this is symmetry. What does symmetry mean? The horizontal distance between A to C is equal to the horizontal distance between C to D. So, the distance X is equal to 2 Xmax. And the distance T is equal to 2 Tmax.

So the time needed from A to C is equal to the time needed from C to D. T to the farthest is equal to 2 Tmax and X to the farthest is equal to 2 Xmax.

Tmax is the formula for V0 sin alpha per G, which means if T is the farthest, 2Tmax is 2V0 sin alpha per G. And the farthest X is 2Xmax. If Xmax is V0 squared sin 2 alpha per 2G, it means if 2Xmax or the farthest X is V0 squared sin 2 alpha per 2G multiplied by 2, the two are separated, so we only have V0 squared sin 2 alpha per G.

Or remember the formula from before? sin2α = 2sinαcosα, right? So we can divide this into V0² + 2sinαcosα, we put the two in front. So, X² = 2V0² sinαcosα/2.

But remember, this special formula only works if it starts from the ground. If it doesn't start from the ground, then we have to use the principle. The principle is this, the position coordinate, which is y = 0, put it into the formula y, y = 0, then the t is the ground t. Then the ground t we put into the formula x, find the x farthest. Okay, understand up to here?

This is a summary of what we have learned earlier. So there is the initial position, point A, certain point B, then this is the highest point, the peak point or C, and the last point D, the point on the ground. So the speed is like this, the position coordinates are like this, and there is a special formula for the peak and ground. But if it's on the ground, this formula only applies to the motion of the ball that starts from the ground. If not from the ground, you have to use the principle.

Okay, understand? Okay, that's it for this video. To see the complete playlist of this pub, you can click the thumbnail on the right hand side. If you have any questions, suggestions, or criticisms, you can write them in the comments column. Hopefully it's useful and see you in the next video.