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Parabolic Path: Definition & Projectiles

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  • 0:02 Definition
  • 0:53 Gravity
  • 1:26 Initial Velocity
  • 2:05 Putting It All Together
  • 6:46 Lesson Summary
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Lesson Transcript
Instructor: Glenda Boozer
The water in a fountain, a human cannonball, or an artillery shell can follow a parabolic path. We'll define a parabolic path and then work our way through the math that tells us where the projectile will be at any particular moment.

Definition

What do the water in a fountain, a human cannonball, and an artillery shell all have in common? Mathematically, they all follow a parabolic path, which is the path followed by a projectile. If you are wondering, a projectile is an object that is only acted on by gravity. In math, a parabola is a curve that is the graphical representation of a quadratic equation.

A projectile has an initial velocity, which is either horizontal or upward at an angle. If a projectile were to start out going straight up, it would go straight down, too. That wouldn't be much of a parabola since it would be squished into a straight line. It would be the same way if it started out going straight down, or if we simply dropped it, so we'll really be looking at the parabolic paths followed by projectiles with some initial velocity, or starting horizontal speed.

Gravity

Gravity gives every object in the vicinity of Earth an acceleration, or increase in speed, of about 32 feet per second every second, or 32 ft/sec^2, or in metric units, about 9.81 m/sec^2. This means that if I drop an object, it will speed up as it falls, ending up at 32 feet per second at the end of the first second, 64 feet per second in the second second, and so on. If we are tracking the position of a projectile, gravity has to be taken into consideration.

Initial Velocity

As we mentioned a moment ago, we'll only be considering situations where the projectile has some initial velocity. This means that it starts out moving, either horizontally or at an angle. If the initial velocity is strictly horizontal, then we only need to know about where the projectile started (maybe ground level, maybe very high up), how fast it starts out going, and how long it has been since it was launched.

If the projectile is aimed upward, then we can think of it as having a vertical component and a horizontal component. It turns out that this, plus gravity, will account for all of the motion, and the starting point will account for the position at any particular moment.

Putting It All Together

This picture shows how the two components of the initial velocity and the constant force of gravity act on a projectile. See the arrows? The diagonal one marked v is broken down into the v sub x and v sub y components. Let's only look at the one at the beginning right now; the others combine the effects of initial velocity and gravity.

parabolic path components

Now let's do the math. We can set up one equation for the horizontal direction, x, and another one for the vertical direction, y. Either one of them will depend on the amount of time since the projectile was launched - let's call it t and measure it in seconds.

The horizontal motion is only influenced by the v sub x component, and of course, by where it starts out. If we use trigonometry, we can find it from v and the angle shown as theta. Since it stays the same as long as the projectile moves (let's not worry about drag right now), every second will see it v sub x further along. This gives us the following:

x = x sub 0 + (v sub x) * t

The vertical motion due to v sub y is almost as simple, so let's take care of that arrow first. It works out just the same as it does for the x. Of course, we start with the original position, y sub 0:

y = (v sub y) * t + y sub 0. Note that v sub y will be positive if the projectile starts out heading upward and negative if it starts out heading downward.

Now all that's left is gravity, but it is always the same for every projectile (on Earth, at least), as we discussed earlier. Since the projectile keeps speeding up as it heads downward (that's negative, remember), the distance piles up according to the square of the time. In feet, it turns out to be:

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