### Video Transcript

In this video, we will look at
projectiles. Specifically, we will analyze the
characteristics of the motion of projectiles as well as the forces acting on
them. To understand what a projectile is,
we must first start by defining the term.

A projectile is an object upon
which only the force of gravity acts. An outside force might be applied
to our object to set it into motion. Projectiles can be dropped, thrown,
flung, or any other manner that would begin their motion. But once they’re in the air, the
only force acting on them is the force of gravity. In the real world, our projectile
feel a resistive force due to air resistance. However, for the purpose of
simplification in our lesson, we will neglect any forces that resist our motion.

With only the force of gravity
acting on our projectile, our object will have a constant acceleration, which will
be the acceleration due to gravity, or little 𝑔. Let’s dig deeper into the motion of
a projectile by looking at a coin dropped from a hand. Let’s draw on the four positions of
our coin falling to better understand what is happening with the coin’s motion as it
falls to the ground.

At position one, as the person
releases the coin, the force of gravity pulls the coin downwards, causing the coin
to begin to move in a downward direction, giving the coin a downward velocity of
𝑣. At some time 𝑡 later, our coin
arrives at position two. The force of gravity continues to
pull our coin downward in the same direction as its motion, causing the coin’s
downward velocity to increase. Meaning the coin is falling faster
at position two than it was at position one.

After another time interval of 𝑡,
our coin is at position three. The force of gravity is still
pulling down as the coin is falling, thereby further increasing the downward
velocity of the coin. Meaning that our coin is going
faster at position three than it was at position two and position one. After another time interval 𝑡
passes, our coin is at position four, where it’s just about to hit the ground. The force of gravity still pulls
down while our coin falls, causing the downward velocity to keep increasing. Meaning that our coin is now at the
fastest it’s been throughout its entire fall.

As our coin falls from position one
to position four, its velocity will increase at a constant rate. This is because as we noted
earlier, a projectile has a constant acceleration of 𝑔. We can also see that the distance
the coin travels between each time interval 𝑡 increases as the coin falls.

Let’s take our analysis one step
further and draw some motion graphs for a projectile. We’ll add a bit of complexity by
throwing a ball up in the air with a velocity of 𝑣 rather than dropping a coin. But we won’t add in any horizontal
motion just yet. Let’s draw in our four positions
again for our ball on the way up and on the way back down.

Just like with the coin, each
position occurs after a time interval 𝑡 has passed. We can see that on the way up, the
coin covers less distance over the same time interval, which means it is slowing
down. If we were to draw in the arrows
that represent velocity, we see that they get smaller as we go from position one to
position four. At position one, we’re given our
ball velocity of 𝑣. At position four, when the ball is
at its highest point in trajectory, it has no velocity because the ball stops for an
instant before it comes back down.

Our ball is slowing down whilst
traveling upwards because the force of gravity is pulling down at each of the
positions shown. When force and velocity in opposite
directions, an object’s motion is slowing down. Even though our ball stops for an
instant at position four, it still has the same force of gravity acting on it as it
did at the other three positions. And so it still has a constant
acceleration downwards of 𝑔. This will cause the ball to start
to come back down with increasing speed as the force of gravity is now pointing in
the same direction as the motion.

After each time interval 𝑡, the
ball will have the same magnitude of velocity as it did on the way up at that
position but will now be pointing downwards. When the ball reaches its starting
position, it would now have a velocity of negative 𝑣. If we’re to make a graph of the
velocity versus time and the speed versus time of our ball in the air. Our velocity-versus-time graph,
assuming that a positive velocity means a velocity in the upward direction and a
negative velocity means a velocity in the downward direction towards Earth, would
look something like this.

We should know that the slope of
the line is a constant and that the value of the slope is equal to the acceleration
due to gravity, or negative 𝑔, where the negative represents the downward
direction. We can add the positions of the
ball that we drew on the side of our screen onto our graph. Position one is when the object is
first thrown and will have a maximum velocity of 𝑣. Position one is also where the ball
comes back to its starting position and has the same magnitude but opposite
direction from the initial velocity with the value of negative 𝑣.

Position two is when the velocity
has a smaller magnitude and has a positive value on the way up and negative value as
the ball comes back down. At position three, the ball’s
velocity has decreased even more, with both a positive and a negative value for the
way up and the way down, respectively. The ball temporarily stops at
position four, the highest point of the trajectory as signified by our graph
crossing the time axis.

Moving on from the
velocity-versus-time graph to the speed-versus-time graph, the speed graph looks
like an absolute value graph of the velocity graph. This is because speed is the
magnitude of velocity and will always have a positive value, whereas velocity also
has a direction and can therefore be negative.

Plotting the four positions on our
speed-versus-time graph, we can see that the speed at each of the positions on the
way up is the same as the speed at each of the positions on the way down. Plotting our
acceleration-versus-time graph, we see that the motion of our ball yields an
acceleration-versus-time graph that is a horizontal line of value negative 𝑔. Which means that the ball has a
constant acceleration whose value is negative 𝑔, which matches the slope of our
velocity-versus-time graph.

We need to keep in mind that we
chose positive to be an upward direction and negative to be a downward direction,
meaning that our acceleration is down towards the Earth. When discussing motion, we must
also take into account the displacement and the distance. So let’s draw those graphs
accordingly.

In our displacement-versus-time
graph, we start off with a large velocity or steep slope. We slow down over the first curve
until we reach the top, where we stop for an instant. Then we begin to come back down
towards the starting place. On the downward slope, our velocity
increases as the slope gets deeper, as we approach the time axis or the ground.

Inserting the positions we have
drawn on the side of the screen onto our graph, we can see the symmetry for our ball
traveling upwards and downwards. At each of the positions, the ball
is the same height for the way up as it is on the way back down. Our displacement is always positive
because we’re always moving on the upward side of where we began. Even though our ball comes back
down, it is always above where it was released.

This graph can be a bit confusing
because it looks like the path that a ball takes when we play catch with a
friend. But we need to remember that we
threw the ball straight up in the air and it came right back down. We need to be careful because this
is not the actual path the object takes, but rather a plot of where the ball is in
its upward and downward motion.

The first half of the
distance-versus-time graph is identical to the first half of the
displacement-versus-time graph, as the ball was moving upwards. When the ball stops for an instant
at the top of its trajectory and begins to move downwards, the distance-versus-time
graph continues to increase. This is because distance does not
have a direction. And therefore, we add on the
additional distances as positive values.

As with the
displacement-versus-time graph and the distance-versus-time graph, the initial slope
is steep and gradually becomes shallow as the velocity decreases. After the midpoint, the shallow
slope becomes steep again as the ball’s speed increases before reaching the starting
position.

Now that we have analyzed a
projectile that has only motion in the vertical direction, we can add a step at our
complexity by analyzing a projectile that also has a horizontal component to its
motion. Let’s look at the motion graphs for
a cannon ball being shot out of a cannon horizontally.

If we were to shoot the cannon ball
out horizontally with a velocity of 𝑣, assuming there’s no air resistance, we need
to remember the only force acting on the cannonball after shot out of the cannon is
the force due to gravity. Which will pull straight down to
the ground at each of the positions shown. The force of gravity is a vertical
force. So it will have the same effect on
the vertical velocity of the cannonball as it did in the previous example.

The force of gravity will have no
effect on the horizontal component of our cannonball as they are perpendicular to
each other, which means that the force of gravity is at 90 degrees to the horizontal
component of the velocity. Since there’s no net force in the
horizontal direction on our cannonball after it’s shot out of the cannon, we can say
that the horizontal component of the velocity of the cannonball will be 𝑣 at each
of the positions shown, one through four. Or we can say that the cannonball’s
velocity in the horizontal direction is constant.

The vertical motion for our
cannonball is the same as the downward motion of the ball in the previous problem,
as our cannonball has an initial vertical velocity of zero. So when we draw our motion graphs,
we’re gonna focus on the horizontal component of our motion. The velocity-versus-time and
speed-versus-time graphs for the horizontal component of the cannonball’s velocity
would look something like this.

Keep in mind that we chose for a
positive direction to be to the right and for a negative direction to be to the
left, as we are talking about the horizontal motion. Looking at our velocity-versus-time
graph, we see that we have a horizontal line of value 𝑣 in the positive direction
or to the right. Notice that the slope of the line
is zero, implying that the acceleration of the cannonball on the horizontal
direction is also zero, which we can connect to the fact that there is no net force
acting on the cannonball in the horizontal direction.

Our speed graph is identical to the
velocity graph with a horizontal line of value 𝑣, as the horizontal component of
the velocity is in only one direction, to the right. Plotting our
acceleration-versus-time graph, we have a horizontal line of value zero, indicating
that the acceleration in the horizontal direction for our cannonball is zero. As noted above in the horizontal
velocity-versus-time graph, where we had a slope of zero.

Moving on to our
displacement-versus-time and distance-versus-time graphs, we should notice that the
displacement-versus-time and distance-versus-time graphs are identical, as our
projectile is moving in only one direction horizontally to the right. As noted previously, the velocity
is constant in the horizontal direction, which means that the slope of our
displacement-versus-time graph, which happens to be the velocity, will also be a
constant.

Now, let’s talk about the most
complicated type of projectile, one which is launched at an angle other than zero or
90 degrees to the horizontal. If we were to kick a soccer ball
such that it left the ground with a velocity 𝑣 at roughly an angle of 30 degrees to
the horizontal, it would travel a path like this. The blue arrows represent the
direction of the velocity of the soccer ball at the five positions shown.

At both position one and position
five, our soccer ball has a velocity of 𝑣 at an angle of 30 degrees to the
horizontal. But at position one, our angle is
up and to the right, where at position five, our angle is down and to the right. Drawing in arrows to represent the
vertical velocity of our soccer ball, we can see that they’re identical to the
arrows that we drew for the ball that we threw straight up in the air. On the way up, our arrows get
smaller, showing that our soccer ball is slowing down, until it reaches the top of
its trajectory at position three, where it has no vertical component of its
velocity. And on the way down, the velocity
gets larger, meaning that it’s going faster as it comes back to the ground.

Once again, this is due to the
force of gravity pulling the object down to the ground at each of the positions
shown. On the way up, the soccer ball’s
velocity is pointing upwards, but the force is pulling downwards, causing it to slow
down. On the way down, the force of
gravity is in the same direction as the velocity, causing it to speed up. The horizontal component of the
velocity of the soccer ball is identical to the horizontal component of the velocity
of the cannonball in the previous problem. This is because there is no
horizontal force acting on the soccer ball after it has been kicked. Therefore, we can say the
horizontal component of the velocity is constant. This is the most complicated type
of projectile cause it combines the two problems we have solved previously.

A common misconception that goes
all the way back to the ancient Greeks is that the soccer ball would travel in a
straight diagonal line up to its highest position and then fall straight back down
to the ground, as shown in black. This, however, is not true. As we’ve established earlier, the
soccer ball is given an initial horizontal component to the velocity when it gets
kicked. Therefore, when it reaches its top
position, at position three, it will still be moving to the right with that same
horizontal component of velocity.

Let’s take a look now at an example
question.

An object is set in motion by an
initial force 𝐹 that acts diagonally upward, as shown in the diagram. The object undergoes projectile
motion. Which of the graphs (a), (b), (c),
and (d) shows the changes in the horizontal displacement of the object between
leaving the ground and returning to the ground?

In the diagram, a force is applied
to an object at an angle close to 90 degrees. The diagram goes on to show the
path that our object takes as it becomes a projectile, goes up to its highest point
and then back down to the ground. Our answers choices are given to us
as horizontal displacement versus time graphs. With graph (a) being an 𝑠 curve,
graph (b) being an 𝑠 curve, graph (c) being a diagonal line, and graph (d) being a
horizontal line.

In the problem, we’re told that our
object undergoes projectile motion. A projectile is an object upon
which only the force of gravity acts. Therefore, we can draw in the force
of gravity acting on our object in the diagram. Our problem asks us about
horizontal displacement. The force 𝐹 is only applied to the
object it throwed into the air. Once 𝐹 is no longer applied, the
only force acting on our object for the rest of the time it’s in the air is the
force of gravity, which acts in the vertical direction. This means that we can say the net
force acting horizontally is going to be equal to zero.

To choose the right
horizontal-displacement-versus-time graph, we need to find a connection between
force and motion. This should remind us of Newton’s
second law, or 𝐹 net equals 𝑚𝑎. The net force acting on an object
is equal to the mass of the object times the object’s acceleration. We just stated that the net force
acting on our object once it’s in the air is zero. So if the net force in the
horizontal direction is zero, that means that the acceleration in the horizontal
direction must also be zero.

Let’s remember that our
acceleration is defined as the change in velocity, or Δ𝑣, over the change in time,
or Δ𝑡. But our acceleration is zero, which
means that there’s no change in velocity of our projectile horizontally while it’s
in the air. Or another way to put it, the
velocity in the horizontal direction is constant.

Recalling our motion graphs, we
should remember that the slope of a displacement-versus-time graph is the
velocity. Therefore, if the velocity is
constant, that means we’re looking for a horizontal-displacement-versus-time graph
with a constant slope. Looking at our answer choices, we
can eliminate answer choices (a) and (b) as they don’t have constant slopes. Also, we can eliminate answer
choice (d). Even though it has a constant
slope, the slope happens to be zero as shown by our horizontal line, which would
imply that we have no horizontal velocity. But as we can see from our diagram,
our ball does have a horizontal component to its velocity as it ends up to the right
of where it started.

Therefore, we can say that graph
(c) would be the correct answer. The graph that shows the horizontal
displacement of the object between leaving the ground and returning to the ground is
graph (c).

In summary for our lesson, a
projectile is an object upon which only the force of gravity acts. The horizontal component of the
velocity is constant with no component of acceleration. The only acceleration is the
acceleration due to gravity, which acts vertically to change the vertical velocity
of the object.