# Physics! #20: University Physics 3.3

Projectile Motion

BOOSH! Okay, so there’s a decent amount of math in this section, but it all relies on stuff you already know. You should make a good effort to understand it, but once again I’m going to try to give you a grip on it, rather than working out all the steps.

We define projectile to mean essentially this: on object given some velocity such that it moves away from them ground only to be pulled back down by gravity. In this case, we don’t care about drag from wind, so we’re talking about very nice parabolic trajectories.

Things like cannonballs, bullets, and catapulted kittens follow parabolic trajectories.

Now, here’s how I like to think about it. When you throw that kitten in the air, you’re giving it two velocity vectors in the Cartesian system: It has some upward velocity and some forward velocity.

Because gravity accelerates the kitten downward, you know the kitten’s upward velocity is going to start at its maximum, slow down, stop, then turn negative as it speeds toward Earth.

Because there is no drag in our idealization, the kitten’s forward velocity will remain the same forever unless something stops it. Something like, say, smacking into the Earth.

So, here’s the cool part. If you know it’s initial starting point, speed, direction, and acceleration, you can where it is at any time, AND when it’ll hit the ground.

This is where vectors make everything nice.

So, we’ll start by telling you that you’re flinging a cat at a certain speed in a certain direction, given by the angle of initial velocity with respect to the surface of the Earth.

Instead of trying to figure out how the ball changes at each point, you first figure how its up-down velocity works. So, using what you know about sines and cosines, you decompose the vector into y and x components.

You know from earlier that there is a very nice equation for position when you have constant acceleration: $y(t) = y_0 + v_yt + \frac{1}{2}at^2$ You want to know when the kitten hits the ground. Well, there are two times when the kitten is at zero altitude: before you fling it, and when it lands. So, we set the position (y) to zero. We also know the acceleration to be g (~10), since you’re on the surface of Earth.

What ho? Is that a quadratic equation? How convenient, since we know there are two times when the cat is at 0 altitude and every quadratic equation has two solutions. One is trivial, since at t=0, the equation zeroes out. The other you solve for by knowing the velocity in the y direction. Thus you know how long the kitten will be flying.

MEANWHILE, you look at your equation for x position. $x(t) x_0 + v_xt + \frac{1}{2}at^2$. You have the same equation as last time, but you can do different stuff to it. Mainly, you know there is no acceleration acting on the kitten. If there were drag in our idealization, you’d have to add that. But, for the moment, just set acceleration to zero. This, and the fact that we define you to start at x=0 leaves you with a nice equation: $x(t) = v_xt.$ Well, you already know t and you already know v in the x direction (because we gave it to you at the beginning). Thus, you solve for the x position when the kitten lands. Thus, you know how far the kitten traveled.

So, essentially the height equation constrains the amount of time the motion takes and the length equation is constrained by time.

Pretty cool, right? You may also notice that the math works just as well for a kitten or a cannonball. This is a simple consequence of what we showed earlier – when you drop something, the time it takes to reach the ground doesn’t depend on its mass.

You may also ask yourself how can I fling the kitten so that it goes as far as possible? First, let’s look at extreme cases.

If you throw the cat directly upward, you’ll have all your velocity in the y direction. So, you’ll get the largest t. However, since you have no x velocity, the cat won’t go anywhere in the x direction. It’ll just land on your head, probably kinda pissed off.

On the other hand, if you throw the cat directly forward, it’ll also go nowhere, since it starts at altitude zero and immediately hits the ground. The kitten is still pissed, but at least is not hurtling toward you from above.

Now, you can visualize a few trajectories. If you throw it a just a little forward of vertical it’ll make a tall thin parabola. If you throw it just a little up from horizontal it’ll make a short narrow parabola.

I’m not going to offer the answer to the ideal angle here, since it’s an example in the book you should work (3.8). So, check it out.

Next up: Motion in a Circle.

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