Video Transcript
In this video, our topic is
vertical motion under gravity. We’re going to learn how to
calculate object motion under the influence of this force and, in doing so, put into
practice a set of equations that help us analyze many real-world scenarios.
When we talk about vertical motion
under gravity, we know that this refers to objects that move only in a vertical
plane, either straight up or straight down. And the fact that we’re talking
about this motion under gravity means that this is the only force acting on the
objects of interest. For this lesson then, we won’t
consider the impact of forces like air resistance or friction or any others.
Something interesting about the
force of gravity is that it causes vertically moving objects to accelerate. For example, when this ball was in
our hand, it had a velocity of zero. But as soon as we let go of it, it
began to accelerate downward. And this means that as time passes,
its downward speed increases. This is why if we were to take
snapshots of the ball at even time intervals, we wouldn’t see the same distance
between the ball each time. It’s falling faster and faster.
To understand vertical motion under
gravity, we’ll need to know a bit about the acceleration that gravity causes, often
represented by a lowercase 𝑔. Near the surface of the Earth, the
magnitude of this acceleration is typically approximated as 9.8 meters per second
squared. Another way of saying this is 9.8
meters per second every second, meaning that for every second that passes for a
freely falling object like our ball here, thanks to the acceleration due to gravity,
the speed of that object will increase by 9.8 meters per second.
This scene of an object in freefall
is one of two types of vertical motion under gravity. Say that rather than dropping our
ball from a certain height, we toss it directly upward into the air. From experience, we know that this
ball will rise to a certain point and then come to rest for just an instant and then
begin to fall back down to Earth, just like the ball we released from rest over
here. We can describe both of these kinds
of motion using what are called the equations of motion. There are four such equations, each
one describing object motion in a different way. All of these equations of motion,
sometimes also called kinematic equations or SUVAT equations, assume an object
moving under a constant acceleration. When that condition is met, these
four equations apply, and when it’s not, they don’t.
Here, we’ve used 𝑠 to represent
displacement, 𝑣 sub 𝑖 to represent initial velocity, 𝑣 sub 𝑓 to represent final
velocity, 𝑡 to represent time, and 𝑎 to represent our uniform acceleration. And in the case of motion under
gravity, our acceleration 𝑎 will always be the same. It’s the acceleration due to
gravity.
Something important to recognize
about these equations is that they’re vector equations. That is, displacement,
acceleration, and velocity are all vector quantities with both magnitude and
direction. So in any situation where we want
to use an equation of motion, it’s important to set up a convention for what motion
is positive and what is negative. In the case of vertically moving
objects, we might decide that any motion upward is positive, meaning that downward
motion is negative. If we decide this way, then all of
our vector variables — displacement, velocity, and acceleration — need to reflect
this convention. For example, since gravity always
points downward, our acceleration due to gravity would be negative 9.8 meters per
second squared. And then over here for the ball
that we dropped from rest, its velocity any moment after that would also be
negative, whereas this ball, which we first tossed up into the air, would have a
positive velocity until it came to rest and then, as it fell back down, would have a
negative one.
Working consistently within a sign
convention is a very important aspect of accurately applying an equation of
motion. Knowing all this, let’s now get
some practice analyzing vertical motion under the force of gravity.
A particle was projected vertically
upward from the ground. Given that the maximum height the
particle reached was 62.5 meters, find the velocity at which it was projected. Take the acceleration due to
gravity 𝑔 to equal 9.8 meters per second squared.
Alright, let’s say that this is
ground level and that this is our particle that’s launched vertically upward. Thanks to that projection, we’re
told that our particle reaches a maximum height, we’ll call it ℎ, of 62.5
meters. What we want to do is figure out
this particle’s initial velocity which allowed it to reach this maximum height
ℎ. So basically, our particle has this
initial upward velocity but is constantly under the downward accelerating influence
of gravity so that as it rises, it moves more and more slowly until eventually it
stops. Since this is an example of an
object accelerating uniformly, in this case, under the influence of gravity, we can
use an equation of motion to describe its motion.
Specifically, we’ll use the
equation that tells us that an object’s final velocity squared is equal to its
initial velocity squared plus two times its acceleration times its displacement. Applying this relationship to our
scenario, we can say that 𝑣 𝑓 squared equals 𝑣 𝑖 squared, where 𝑣 𝑖 is what we
want to solve for, plus two times 𝑔, that’s the acceleration our particle
experiences, multiplied by ℎ, the maximum height it attains. Recall we mentioned earlier that
when a particle’s at its maximum height, its speed is zero. So if we let this moment in time be
the final one, represented by our subscript 𝑓, then that tells us 𝑣 sub 𝑓 is
zero. And so this whole term is zero. So then this is the equation we
want to solve for 𝑣 sub 𝑖.
Now at this point, let’s set up a
sign convention so we can say what kind of motion in this situation is positive and
what is negative. Let’s say that 𝑣 sub 𝑖, the
initial velocity of our particle, is a positive quantity, which means that any
vector pointed in the opposite direction will have a negative sign. And we see right away that the
acceleration due to gravity points downward. So when we substitute in for 𝑔 in
this equation, we’ll use negative 9.8 meters per second squared.
Let’s perform that substitution now
along with replacing ℎ by 62.5 meters. And this quantity is positive
because ℎ is vertically upward from the ground. Leaving out the units for now, we
have zero equals 𝑣 sub 𝑖 squared plus two times negative 9.8 times 62.5. And then if we add the product of
these three quantities to both sides of our equation, we find that 𝑣 sub 𝑖 squared
equals two times 9.8 times 62.5. And then taking the square root of
both sides, we see that 𝑣 sub 𝑖 is given by this expression, which comes out to 35
with units of meters per second. So this is the velocity at which
our particle was projected.
Now let’s look at an example
involving an object dropped from rest.
If a body, which was dropped from a
building, took three seconds to reach the ground, find its average velocity as it
fell. Let the acceleration due to gravity
𝑔 equal 9.8 meters per second squared.
Alright, so let’s say that this is
our building. And here we have our body, this
pink object, that is dropped vertically downward from the building. Say the person who dropped it
starts a stopwatch as soon as the object is released. We’re told that it takes three
seconds for the object to reach the ground. Knowing all this, we want to solve
for the average velocity of this body as it falls. Now, because this body is under the
influence of no force other than gravity, we know that it accelerates uniformly. This means we can use what are
called the equations of motion to describe our falling body. Specifically, we’ll use the
equation that tells us that an object’s final velocity is equal to its initial
velocity plus its acceleration times the time over which it was accelerating.
Along with this, we’ll use the fact
that for a uniformly accelerating object, the average velocity of that object equals
the sum of its final and initial velocities divided by two. That average velocity is exactly
what we want to solve for here. And to figure it out, we’ll need to
know our body’s final as well as initial velocities. We know that our body was dropped,
that is, released from rest, which tells us that it had an initial velocity of
zero. And that means to solve for its
average velocity, we just need to know its final and then divide that by two. And this is where we can use the
fact that that final velocity equals initial velocity plus 𝑎 times 𝑡.
And since we’ve seen that 𝑣 sub 𝑖
is equal to zero and that in this case our acceleration is the acceleration due to
gravity 𝑔, we can write that 𝑣 sub 𝑓 is equal to 𝑔 times 𝑡. And we’re told that that time 𝑡 is
three seconds and that the acceleration due to gravity 𝑔 is 9.8 meters per second
squared. If we substitute in these values
leaving out the units, we have that 𝑣 sub 𝑓 is equal to 9.8 times three or 29.4
with units of meters per second. We can now take this value and
substitute it in for 𝑣 sub 𝑓 in our equation for 𝑣 sub average. So we find that 𝑣 sub avg equals
29.4 over two or 14.7 meters per second. That’s the average velocity of our
body as it falls.
Let’s look now at another example
exercise.
A particle is projected vertically
upward at seven meters per second from a point 38.7 meters above the ground. Find the maximum height the
particle can reach. Consider the acceleration due to
gravity to be 9.8 meters per second squared.
Alright, so if we say that this is
ground level, then if we go 38.7 meters up from there, we have our particle where it
is projected with an upward velocity of seven meters per second. This tells us that our particle
will continue moving upward. But we know that under the
influence of gravity, it will slow down more and more until eventually it comes to
rest. At this point, it has reached its
maximum height. We’ll call it ℎ sub max. This is the height we want to solve
for, and we can see it’s equal to 38.7 meters plus this height here. We’ll call this height ℎ. And since as our particle moves
across this height it’s accelerating uniformly under the influence of gravity, we
can use an equation of motion to solve for ℎ. Specifically, we’ll use the
relationship that an object’s final velocity squared is equal to its initial
velocity squared plus two times its acceleration times its displacement.
Applying this relationship to our
situation, we’ll say 𝑣 sub 𝑓 squared equals 𝑣 sub 𝑖 squared plus two times 𝑔,
that’s our particle’s acceleration, multiplied by ℎ. Going back over to our sketch, if
we say that our particle’s position here is its initial position, where it had an
initial velocity 𝑣 sub 𝑖, and our particle’s position here at its maximum height
is its final position, then we can say that its final velocity 𝑣 sub 𝑓 is zero,
meaning that this is then the equation we have to solve for the height ℎ.
At this point, we can set up a sign
convention where we say that motion upward is positive and motion downward is
therefore negative. This is useful to us because the
acceleration due to gravity is downward, while our initial velocity, what we’ve
called 𝑣 sub 𝑖, is upward. And so leaving off the units, this
means that we would use a value of positive seven for 𝑣 sub 𝑖 and negative 9.8 for
𝑔. Substituting in these values, if we
then subtract seven squared from both sides of this equation, we have that negative
seven squared is equal to two times negative 9.8 times ℎ. And both negative signs drop
out. And if we then divide both sides of
the equation by two times 9.8, we find that ℎ is equal to 49, that’s seven squared,
divided by two times 9.8. That’s equal to 2.5. And we’ll include the units of
meters.
Recall though that this isn’t our
answer because ℎ is just one part of ℎ max. ℎ max is equal to 38.7 meters plus
ℎ or 38.7 meters plus 2.5 meters. And adding these together, we get
41.2 meters. This is the maximum height our
particle can reach.
Let’s look now at another example
involving vertically projected motion.
If a body is projected vertically
upwards with speed 𝑉 to reach maximum height ℎ, then the speed the body should be
projected by to reach height four ℎ is blank. (A) 𝑉, (B) four 𝑉, (C) two 𝑉,
(D) square root of two 𝑉.
Alright, so here we have a scenario
where a body is projected vertically upward with a speed we’ve called 𝑉. And under this influence, it
achieves a maximum height ℎ. We then imagine a scenario where
this same body is projected to a height of four ℎ. And the question is, with what
initial speed would it need to be projected to reach this height? We have these four answer options
here. And as we get started with our
answer, we can notice the fact that this body, as it moves upward, is under the
influence of only the force of gravity. Therefore, its acceleration is
uniform, and we can describe its motion using an equation of motion. The equation we’ll use is that an
object’s final velocity squared is equal to its initial velocity squared plus two
times its acceleration times its displacement.
In the case of our vertically
projected bodies, we can say that the final moment in time is the one where each one
is at its maximum height. And in each case at this point, its
velocity is zero. This means that the left-hand side
of this expression will be zero. And then, as we fill in the
right-hand side, let’s focus on this case where we have a maximum height ℎ and an
initial speed 𝑉. We would write then that zero is
equal to 𝑉 squared plus two times 𝑔, the acceleration our body undergoes,
multiplied by ℎ. It’s possible to rearrange this
equation so that it reads ℎ is equal to negative 𝑉 squared over two times 𝑔.
And in this instance, 𝑔 is equal
to negative 9.8 meters per second squared. And writing that value in without
units, we see that the negative signs in numerator and denominator cancel. So if we want a body to ascend to a
maximum height of ℎ, we need to give it an initial speed of 𝑉. And if we want a body to ascend to
a maximum height of four ℎ, where we multiply both sides of the equation by four,
then we can equivalently say that this is equal to two times 𝑉 quantity squared
divided by two times 9.8. And so, now we know what our
object’s initial velocity must be in order for it to be projected to a height of
four times ℎ. It must be twice the initial speed
𝑉.
Let’s finish up now by reviewing a
few key points from this lesson. First, we’ve seen that a body
moving vertically under gravity is described by the equations of motion. To solve for quantities like
displacement, velocity, time, and so on, making a sketch and establishing sign
conventions is often helpful. And lastly, memorizing the
equations of motion can be helpful too.
