Physics IV: Projectile Motion

Community

Physics

Physics IV: Projectile Motion

Author: oLahav

This series of lessons is designed to help you learn, or review, the fundamentals of physics. We now combine acceleration with speed and vector addition in a concept called projectile motion.

Projectile what exactly?

Projectile motion.

Things that move in a trajectory. For example, a ball kicked in a parabola:

As a matter of fact, anything that moves in a parabola with no external forces other than gravity is considered to be in projectile motion.

So what do we do with this?

The key thing about projectile motion- there are 2 components: horizontal, and vertical.

The horizontal.component is the easy one. There's no horizontal gravity (remember, gravity only pulls down), or acceleration of any kind, so we're dealing with constant speed:

$v_h=\frac{d_h}{t}$ .

The vertical component does have gravity, so it's a bit more complicated. We're dealing with equations like

$a=-9.8=\frac{v_{v_f}-v_{v_i}}{t}$ .

These two components have exactly 1 thing in common- time. Everything else- speed, distance, acceleration- is different.

What do these questions look like?

The questions can have many variations, but they almost always ask you to figure one component out using the other. You'll always be given enough information, but you may have to add a few things on your on, such as:

The speed at the top of the projectile is always 0. If the projectile lands at the same height as where it started from, this will also represent half the time.

The speed at start and at impact is NOT 0, this may be confusing.

Gravity is constant acceleration measuring

$9.8\frac{m}{s ^2}$

and is always negative.

An example can clear things up:

Example:

I kick a football at a speed of 10sqrt2 m/s at 45 degrees off the ground. How far will the ball travel horizontally before it hits the ground?

We need to find the horizontal distance. First, let's separate our components. Look at this triangle:

The hypotenuse represents the speed we are given, but using the angle we can separate this into horizontal and vertical components. These are the 2 speed vectors that add up to the original one, so that's where vectors come into play.

Now, we know the horizontal speed, and we're looking for the horizontal distance. But we still need time! What do we do?

Luckily, time is the same both horizontally and vertically, so we can use our information on vertical components to find it, and apply it horizontally. We know that at half the time, the top of the parabola, the speed was 0, and we know initial speed and acceleration. So:

$-9.8=\frac{0-10}{t}$ , or t=1.02 s

Great, and now plugging into the horizontal section, we get

$10=\frac{d_h}{1.02}$ , so d_h=10.2

Therefore, we have just applied the vertical component to find that the horizontal distance was 10.2 m. That's our answer.

Of course, questions aren't always exactly like this one- they can come in many shapes and variations. But this is the basic form- using one component to find information on the other and then solve the problem.

Projectile motion is a great application of what we've learned so far

Next time, we'll get into a whole new aspect of physics- forces.

Thanks for reading this Welcome to Physics Lesson!