Video Transcript
In this video, we’re going to learn
about projectile motion. We’ll learn what it is. We’ll understand the equations that
describe this type of motion. And we’ll also see how this motion
works through a few examples.
As we get started, imagine that you
and your family are on summer vacation at a lake. It’s a lazy sunny afternoon. And you’re relaxing on the dock
with your water balloon launcher and a stack of already filled-up and tied-off water
balloons nearby. Just as you’re about to nod off,
you notice, out of the corner of your eye, one of your siblings paddle in a canoe
out on the lake. Slowly the canoe moves along. And an idea comes to mind. What if you were to load up one of
your water balloons into the launcher and try to land it inside the moving
canoe. In order to know how far back to
pull the launcher and at what angle we can launch the balloon, we’ll want to know
something about projectile motion.
An object experiences projectile
motion when it moves to the air influenced by no other force than the force of
gravity. Examples of this motion could be a
rock falling straight down, a basketball arching towards the basket, or an arrow
fired from a bow. One helpful fact to know about
projectile motion is that motion in the horizontal direction and motion in the
vertical direction is independent of one another. For example, in the case of the
basketball, we could analyze its horizontal motion. And that analysis would be
completely separate from its vertical motion, and vice versa. We know that objects being affected
only by the force of gravity will move. But the question is, how did they
move. That is, how do we describe their
motion. It’s the kinematic equations which
describe the motion of the uniformly accelerating objects that we can look to, to
help us out with projectile motion.
These four equations help us
understand and solve for the motion of projectiles. Using them, we can solve for terms
such as an object’s initial velocity, 𝑣 sub zero, or its displacement, 𝑑, or the
time a certain process takes, 𝑡. In these four equations, there are
no forces involved. And all of them are based on the
assumption that acceleration 𝑎 is constant. Since projectiles, by definition,
experience no other acceleration than the acceleration due to gravity, 𝑔, there is
a natural fit between the kinematic equations and projectile motion. We will use these kinematic
equations to practice some examples in a minute. But before we do, one last note
about projectile motion. And that is, that signs, plus or
minus, are very important when we solve kinematic equations or projectile motion
exercises. Given a two-dimensional projectile
motion problem, it’s very important to decide which directions are positive. And which are negative. That will affect the signs of the
various values we use in the kinematic equations. With that said, let’s try a few
examples of projectile motion exercises.
A crossbow is aimed horizontally at
a target 27 meters away. The crossbow’s bolt hits the target
12 centimeters below the spot at which it is aimed. What is the initial horizontal
velocity of the bolt?
Labelling the distance between the
crossbow and the target, of 27 meters, as 𝑑 and calling the distance below the
horizontal line, 12 centimeters, ℎ. We want to solve for the initial
horizontal velocity that the bolt has. We’ll call that 𝑣 sub 𝑥. To begin on our solution, let’s
draw a diagram of this process. Our scenario begins with a crossbow
aimed horizontally a distance 𝑑, 27 meters, away from the target at which it
aims. But, after the bolt is fired and we
look at where it actually ends up on the target, we see it’s a distance ℎ of 12
centimeters below that horizontal line. We want to find out, when the bolt
initially left the crossbow, what was its horizontal velocity. We can recognize that when the bolt
is in the air, the only force acting on it is the force of gravity. It’s being constantly accelerated
by that force. And therefore, its path is an
example of projectile motion described by the kinematic equations.
Before we look through this set of
equations to find which one might help us the most to solve this exercise, let’s
look back at our diagram for a moment. If we define a set of coordinate
axes, 𝑦 for vertical and 𝑥 for horizontal, with the origin at the place where the
bolt was released by the crossbow. Then we can choose to define up,
vertically, as motion in the positive direction and motion to the right as positive,
in the horizontal direction. By this convention, our height, ℎ,
changes from positive to negative 12 centimeters. And when we define the value of the
acceleration, 𝑔, due to gravity, due to our sign decision, that value will be
negative 9.8 meters per second squared. In order to solve for 𝑣 sub 𝑥, if
we knew how much time, 𝑡, the bolt is in the air. We could figure out how long it’s
taken to travel the horizontal distance 𝑑 and then work back to get 𝑣 sub 𝑥.
All that motion is motion in the
horizontal direction. But to solve for 𝑡, we will want
to look in the vertical direction, at the bolt as it falls due to gravity. So focusing on that axis, the
vertical axis, let’s revisit our kinematic equations and see if we find one that can
help us solve for 𝑡. Given the information we know about
motion in the vertical direction, that objects accelerate according to 𝑔 and that
our displacement is negative 12 centimeters. There isn’t a single equation we
can use to solve for 𝑡. But we can move in that
direction. If we choose the second kinematic
equation listed, using our variables. We could write that 𝑣 sub 𝑓, the
final vertical velocity of the bolt, squared is equal to its initial vertical
velocity squared plus two times the acceleration due to gravity times ℎ. We’re told that the bolt is fired
horizontally. So its initial vertical velocity is
zero. Solving for its final vertical
velocity then, we can take the square root of both sides. Since we know both 𝑔 and ℎ, the
value 𝑣 sub 𝑓 is one we can consider known.
Now, back to our list of kinematic
equations. If we know 𝑣 sub 𝑓, the final
vertical velocity of the bolt, then what equation can we use now to solve for the
time 𝑡. Looking over the list, we see that
the first equation written can be helpful. Writing this equation in terms of
our variables, it says 𝑣 sub 𝑓 is equal to 𝑣 sub zero, with these velocities
again referring to motion in the vertical direction, plus 𝑔 times 𝑡. Just like before, 𝑣 sub zero is
equal to zero since the bolt moves horizontally at first. And if we then divide both sides by
the acceleration due to gravity, 𝑔, we find that 𝑡, the time that the bolt is in
the air, is equal to its final vertical velocity divided by 𝑔. If we plug in our expression for 𝑣
sub 𝑓, we see this simplifies a bit. Simplifying to the square root of
two times ℎ divided by 𝑔.
We’ve now effectively solved for
the time the bolt is in the air, by considering only motion in the vertical
direction. We’ll now switch over to motion in
the horizontal direction. We can recall that average velocity
is equal to distance travelled divided by time it takes to travel that distance. 𝑣 sub 𝑥, the bolt’s initial
horizontal speed, is equal to the distance it travels horizontally, 𝑑, divided by
the time it takes to travel that distance. If we plug in our expression for 𝑡
and then simplify the resulting expression, we find that 𝑣 sub 𝑥 is equal to the
square root of 𝑔 over two ℎ all multiplied by 𝑑. We know the values for 𝑑, ℎ, and
𝑔 and can plug them in now. When we do, we’re careful to
convert our height from units of centimeters to units of meters. So that it’s consistent with the
rest of the values in this expression. When we enter these terms on our
calculator, we find that, to two significant figures, 𝑣 sub 𝑥 is 170 meters per
second. That’s the initial horizontal
velocity of the bolt.
Now, let’s try another example
involving motion solely in the vertical direction.
An artillery shell is fired
vertically upward at a target 200 meters vertically above the ground. When the shell is 100 meters above
the ground, it has a speed of 100 meters per second. What is the speed of the shell when
it hits the target? Neglect drag forces on the
shell.
We can name the speed that the
shell has when it hits the target 𝑣 sub 𝑇 and begin in our solution by drawing a
diagram of this process. In this example, we have a
vertically pointed artillery canon firing a shell straight up. When the shell passes the 100-meter
elevation mark, it’s moving at 100 meters per second. We want to know, when it reaches
its target at 200 meters, what is its velocity, 𝑣 sub 𝑇. In this scenario, we can define
vertical motion as motion beginning at ground level and positive in the upward
direction. When the shell leaves the canon, it
will be subject to only one force, the force of gravity acting down on it. Because of our sign convention, we
write 𝑔 as negative 9.8 meters per second squared. Since the artillery shell is
accelerating constantly, due only to the effects of 𝑔, we know that the kinematic
equations apply to this scenario.
Looking over these equations, in
terms of the variables they’re written in, we want to solve for a final velocity 𝑣
sub 𝑓. That’s what 𝑣 sub 𝑇
represents. Both of the first two equations
written show that variable. So let’s see if there’s more
information that can help us choose between these two. Looking at our diagram, based on
our problem statement, we see that we know distances in the vertical direction. And we also know the acceleration
acting on the shell. When it comes to the variables
involved, the only difference between the first and the second kinematic equation
have to do with time versus distance. Since we’re given distance, but not
time, let’s choose to work with the second equation.
Writing it in terms of our
variables, we can say that 𝑣 sub 𝑇 squared is equal to 𝑣, the velocity of the
shell when it’s at 100 meters elevation, plus two times 𝑔 times 𝑑. Where 𝑑 is the distance between
that 100- and 200-meter mark. Taking the square root of both
sides of the equation and given 𝑣 and 𝑔 and recognizing that 𝑑 must be equal to
100 meters, we’re ready to plug in and solve for 𝑣 sub 𝑇. With these values plugged in,
notice how important the signs are, that we established a sign convention and
followed that with these values. It’s the sign convention that tells
us that 𝑣 is positive, 𝑔 is negative, and 𝑑 is positive. When we calculate this expression,
we find that, to three significant figures, it’s 89.7 meters per second. That’s the velocity of the shell in
the upward direction when it reaches the target.
Let’s summarize what we’ve learned
about projectile motion so far. Projectile motion occurs when
objects move through the air under the influence only of the force of gravity. Also, we’ve seen that vertical
motion and horizontal motion are independent from one another and are treated
separately. Kinematic equations, the set of
four equations that apply when acceleration is constant, help us solve for the
motion of projectiles. And finally, we saw the importance
of keeping our signs, positive and negative, in order. Applying what we’ve learned, we’re
able to make sense of projectile motion.
