PROJECTILE MOTION INTRODUCTION

A “projectile” is defined as an object subject only to the force of gravity and no other forces (similar to our definition of free fall).  So this eliminates anything self-propelled like rockets and airplanes.  Good examples are a thrown or batted baseball, a thrown or kicked football, an arrow, a bullet, etc.

What shape do we find when we throw, or fire, a ‘projectile’?

It’s a parabola.

What two motions can you break this up into?

it goes up-down, or vertical, motion

it goes forward, or horizontal motion

We can determine what happens in the y-direction separate from what happens in the x-direction.  Why?  The motions are independent of each other: horizontal velocity has no effect on vertical velocity, and vice-versa.

So we need to determine how fast an object is moving upward versus how fast it is moving outward, or horizontally.

Take a ball thrown at 20 m/s at an angle of 60 degrees:

It moves upward, but also moves forward.  How can we determine how much it moves in either direction?

Ideas?

If we draw a right triangle around the 20 m/s velocity arrow:

Can you now think of some trig functions to determine the height and base of the triangle?

If you said sine and cosine, you’re right (if you didn’t, I guess I’ll never know).

To find the VERTICAL VELOCITY we use the sine function:

(remember SOHCAHTOA?)

 Using our numbers:

To find the HORIZONTAL VELOCITY we use the cosine function:

In words this result says:  For every second that goes by, the object moves upward 17.32 meters and outward 10 meters.  Clearly it starts out moving upward faster than it moves horizontally.  This changes with time, however.

The general formulas for these two sides of the triangle are here:

where q = the angle.

If we look at what happens to the different parts of the velocities:

In this drawing, one arrow represents the y-velocity (vy), one the x-velocity (vx), and one the general direction of the object (v).  The length of these arrows is related to how large a value they have.  So a very short arrow means a very small velocity.

What’s happening to the vy arrow as the object climbs in the air?  How about as it falls back down?  Look at the diagram first, then read on.

As it goes up, you should have noticed the vy arrow gets smaller until it reaches zero at the top: just like our free fall examples.  As it comes back down it reverses direction and then becomes larger in the negative direction.

Now what happens to the vx arrow?  Follow it carefully in each step before reading on.

The vx arrow never changes!  That means the x-velocity is unaffected.  In introductory physics (and even in higher level physics) we ignore air resistance.  In real life, the x-velocity would decrease, but it’s too complicated to calculate and hides the basics.  So happily we only have the y-direction to worry about any acceleration (gravity).

We now modify the kinematic equations that you’re all experts in and write them with the following changes:

1. We change “a” to “-g” to represent gravity.  On Earth, g = 9.8 m/s2.  On the Moon, g = 1.63 m/s2,  and so on.
2. We add another subscript to velocity: we add a “y”.
3. We add a second dimension…x !   More about that later.

So,

And we add a fourth equation for the x-direction:

It’s a very simple equation:        distance = rate x time.

Thus we have the projectile motion equations:

Example:

An archer shoots a bow for distance.  The velocity of the arrow is 22 m/s and he fires at an angle of 42o.

a) how long will it take to reach maximum height?

b) how long will it take to come back down?

c) find out how far (Dx) it goes horizontally.

 Dx

When solving these problems, remember these points:

1. always get vix and viy first, even if unasked.
2. solve for the y-motion just as you did with free fall, using the viy number.
3. solve for the distance using the Dx equation (#1) and vix.

first find vix and viy:

= (22 m/s) cos (42 deg) = 16.34 m/s

= (22 m/s) sin (42 deg) = 14.72 m/s

a) Solution:

What 3 things do we know?

1. viy = 14.72 m/s
2. g = 9.8 m/s2
3. at the top of the arc, vfy = 0, just as in our free fall problems.

Now use equation #2 of the projectile motion equations above:

0        =  14.72 m/s  -  (9.8 m/s2)t

That’s the answer to (a).  I’m betting you can guess the answer to (b).

Which brings us to part (c): how far did it go?

For this we use the 1st equation with Dx:

Dx = vix t

Dx = (16.34 m/s) (3.004 seconds)     Notice this is the TOTAL time!!!!

Dx = 49.1 meters         answer.

You’re now on your own…