 Lesson Video: Vertical Motion under Gravity | Nagwa Lesson Video: Vertical Motion under Gravity | Nagwa

# Lesson Video: Vertical Motion under Gravity Mathematics

In this video, we will learn how to use the kinematics equations of uniform acceleration to model the vertical motion of a body with uniform acceleration due to gravity.

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### 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.