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>> In this unit,
we've described motion
dealing with movement
in one dimension.
Up and down, like this.
One line.
But now, we're ready for 2D.
This segment is about
horizontally launched
projectiles that move along
a two dimensional plane.
We've been looking at
how things move in
one dimension, length.
The object can go
in any direction,
but it is only traveling
in a single line,
one direction at a time.
In two dimensions,
you can travel two ways,
such as length and height.
Like when you draw something
on a sheet of paper.
But there's no thickness
or width, which is
the third dimension.
Let's see how a projectile,
which is an object that is
moving through the air,
moves through space
in two dimensions
in a trajectory
along horizontal
and vertical directions.
To understand what happens,
let's set up a problem about
a horizontal projectile
and we'll solve it
later in the video.
I've got a couple of
high school baseball players
to help us answer the question:
which will hit
the ground first?
A ball thrown horizontally
by the pitcher
or a ball dropped by
the catcher at the same time?
We'll get back to that,
but first, let's look at
how a ball moves through
a two dimensional space.
Horizontal and vertical
motions of an object are
independent of one another.
When you throw an object,
once it leaves your hand,
there are no forces
acting on the ball
as it travels horizontally.
So it would continue to move
at a constant velocity forever,
if it wasn't for another force.
When you throw an object,
the force of your hand
only affects the object
while it is still
in contact with you.
Once it leaves your hand,
there are no forces
acting on the object
as it travels horizontally.
So it would continue to move
at a constant velocity forever
if it wasn't for another force.
The ball is traveling
at a constant velocity
horizontally,
but in the vertical direction,
it's a different story.
What force is always
bringing you down
as long as you
are on planet Earth?
That's right, gravity.
Never gives up,
and it constantly
accelerates the ball downward.
Before we go back
and look at the question
we asked about which ball
will hit the ground first,
we need to get a complete
picture of how
each ball behaves.
Vertical displacement is the
missing piece we need to solve
to get that big picture.
Let's look at the equations
you need to do that.
You only need one equation
to solve for an unknown
in the horizontal direction,
which is velocity equals
distance divided by time.
That's because
there's no acceleration
in the horizontal, remember?
In the vertical
direction though,
we are feeling the effects
of gravity constantly
accelerating downward.
So we use the three equations
that have acceleration in them.
Final velocity equals
your initial velocity plus
your acceleration times time.
Your final velocity squared
equals your initial
velocity squared
plus two times
your acceleration
times displacement.
And your displacement equals
your initial velocity
times time plus 1/2
your acceleration
times your time squared.
How do these variables
translate into what we'll
need to solve our problem?
First, let's look at the
trajectory of the ball
when it's thrown
horizontally.
It may look like it's traveling
in a straight line,
but remember gravity.
It's doing its thing here,
accelerating the ball downward.
Remember free fall.
The acceleration due to gravity
is 9.8 meters
per second squared.
Now, the distance that the ball
travels in the X direction,
the horizontal direction,
is the range.
The amount of time that
the ball spins in the air
is called its hang time.
What direction is
the velocity in?
It's all in the X,
the horizontal.
What we want to solve for
is the displacement
in the vertical direction.
That's the unknown.
The same approach can be taken
as for one-dimensional motion.
The same equations for
velocity, acceleration,
and displacement
all still apply.
We just divide the problem
into horizontal
and vertical components
because horizontal
and vertical motion
don't affect each other.
And because horizontal
and vertical motion
are independent,
we split the directions
into X and Y coordinates
when we solve horizontal
projectile problems.
We use the kinematic equations
we already learned,
which we can apply to both
horizontal and vertical.
Remember that our acceleration
in the horizontal direction
is 0.
That's why we'll only use one
equation for that direction.
Let's use two baseballs.
We will have one player drop one
and another player
will throw a ball.
The heights are the same.
When footage of the two balls
is slowed down,
you can see that
they hit the ground
at exactly the same time.
Surprised?
Most people are.
But the balls are just
following the laws of physics.
The path that the two balls
take through the air
is called their trajectory.
Now let's solve a horizontal
projectile problem together.
The standard distance
from pitcher to catcher
in high school baseball
is 18.4 meters.
First, remember to choose
what direction is
positive direction.
Let's make that down
and to the right.
That makes all of
our values positive,
which makes it easier
to solve the problem.
Second, we fill the chart
using the values we are given.
The ball falls a vertical
distance of 1.5 meters.
So D sub Y equals 1.5 meters.
The ball travels a distance
of 18.4 meters in
the horizontal direction.
This is our range.
You might think that you
don't know any other values,
but you do.
Remember the constant for
the acceleration due to the
force of gravity on Earth?
When we are accelerating
in the vertical direction
solely under the influence
of gravity, our acceleration
is that constant.
9.8 meters per second squared.
So vertical acceleration is
9.8 meters per second squared.
Also, in the vertical direction,
what is our initial velocity?
Well, when we start, it's 0.
The ball hasn't started to drop.
But our unknown quantity
is the horizontal velocity
of the ball as it leaves
the pitcher's hand.
So now that we have
all of this information,
we need to solve for time
and use that to get velocity.
How fast the ball
is traveling.
Time is the one variable
that is the same for X and Y.
If we solve for it
in the Y direction,
we can use it
in the X direction.
Now let's solve
for our unknowns.
We have enough information
in the Y direction column
to solve for our hang time.
What equation has time,
acceleration, vertical velocity,
and vertical distance in it?
How about this one?
Let's plug in our variables.
1.5 meters equals
0 meters per second
multiplied by our time
plus half of our acceleration,
9.8 meters per second squared,
multiplied by our time squared.
Our equation now
looks like this.
1.5 meters equals 0 times time
plus 1/2 times 9.8 meters
per second squared
times time squared.
Because this term is
multiplied by 0,
this whole term disappears
because 0 times anything is 0.
So our new equation
looks like this.
1.5 meters equals 1/2 times
9.8 times our time squared.
Half of 9.8 is 4.9,
so our equation now
looks like this.
1.5 meters equals
4.9 times our time squared.
We want to get time alone,
so we will divide both sides
by 4.9 meters
per second squared.
Our equation looks
like this now.
1.5 meters divided by 4.9
equals 4.9 times time squared
divided by 4.9,
where the 4.9's cancel out of
the right side of the equation.
1.5 meters divided by
4.9 meters per second squared
is 0.31, so our equation
now looks like this.
0.31 equals time squared.
To get time alone and get
rid of the squared sign,
we must take the square root
of both sides.
When we do this,
we get our final answer,
0.55 seconds.
Now that we have time
in the Y direction,
we can use that
in the X direction
to solve for our velocity.
The only equation we can use
in the X direction
is V sub X equals
D sub X divided by T.
When we plug in the numbers
we now know,
we find that our velocity
in the X direction
is equal to our D sub X.
18.4 meters divided by
our time, 0.55 seconds.
We get a final answer
of 33.5 meters per second.
That is a pitch thrown
at 75 miles per hour.
Not bad.
Now we've looked at
horizontally launched
projectiles, how they behave,
and how we can calculate
the speed of one.
For a more in depth look
at projectile motion
at an angle, check out
our Closer Look.
That's it for this segment
of "Physics In Motion."
We'll see you next time.
>> For more practice problems,
lab activities,
and note-taking guides,
check out the
"Physics In Motion" tool kit.