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 (v_{y}), one the
x-velocity (v_{x}), 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 v_{y} 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 v _{y} 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 v_{x }arrow? Follow it carefully in each step before
reading on.

*The v _{x }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:

- We change “a” to “-g” to represent
gravity. On Earth, g = 9.8 m/s
^{2}. On the Moon, g = 1.63 m/s^{2}, and so on. - We add another subscript to velocity:
we add a “y”.
- 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 42^{o}.

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:

- always get v
_{ix}and v_{iy}first, even if unasked. - solve for the y-motion just as you did
with free fall, using the v
_{iy}number. - solve for the distance using the Dx
equation (#1) and v
_{ix}.

first find v_{ix}
and v_{iy}:

_{} = (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?

- v
_{iy}= 14.72 m/s - g = 9.8 m/s2
- at the top of the arc, v
_{fy}= 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/s^{2})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 1^{st} equation with Dx:

Dx = v_{ix}
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…