Lesson Video: Projectile Motion | Nagwa Lesson Video: Projectile Motion | Nagwa

Lesson Video: Projectile Motion Physics • First Year of Secondary School

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In this video, we will learn how to analyze the motion of objects that move horizontally while undergoing constant vertical acceleration.

15:29

Video Transcript

In this video, our topic is projectile motion. Specifically, we’ll be learning some lesser known equations that describe this motion. We’ll use these to compute the range of a given projectile, its maximum altitude, and its time-in-flight.

When we talk about projectile motion, we’re speaking of objects that move under the influence of just one force, gravity. So, for example, say that we have a potato launcher that fires a potato through the air. If we neglect air resistance, then we can say that this potato is moving only under the influence of gravity. And so, this potato is a projectile. For one thing, that means that this potato’s motion is described by what are called the equations of motion. There are four such equations, and they describe properties of our projectile, such as its final velocity, its initial velocity, its acceleration, displacement, and time elapsed. In general, the equations of motion can be used on any object that experiences constant acceleration. That is, as long as 𝑎 is always the same, then these equations apply. In the case of our launched potato and any other airborne object, its acceleration is constant, and it’s the acceleration due to gravity 𝑔.

Now, as we consider our potato’s overall trajectory, there are a few different quantities that we might want to calculate. The first thing we might want to know is how far horizontally does our potato travel. This is called the range of our projectile. We might also want to know the maximum altitude or elevation that our projectile achieves. We can refer to this symbolically as ℎ sub max. And lastly, we might want to know the total time that our potato is in the air. We’ll call this its time-in-flight and represent it as 𝑡 sub 𝑓.

It turns out that to solve for these three quantities range, maximum height, and time-in-flight, we can use the equations of motion. In order to do this, though, we’ll need to know something about the initial conditions of our launched projectile. Say that we take a very up-close view of our potato launcher just as it launches this potato. Over this very short time interval, the potato’s trajectory would essentially be a straight line out of the launcher. And we might be told that the potato initially moves with a speed we’ll call 𝑣 sub 𝑖 and that it’s launched at an angle we’ll call 𝜃 above the horizontal. If we know these two things, the initial speed and the initial angle of our projectile, then working with the equations of motion, we have everything we need to solve for the range, the maximum height, and the time-in-flight.

To see this, let’s start by solving for 𝑡 sub 𝑓. Now, according to our first equation of motion, the final velocity of our object, in this case a potato, is equal to its initial velocity plus its acceleration times its time. If we separate out our projectile’s motion into horizontal and vertical components, then we can apply this equation of motion to either dimension, vertical or horizontal. Let’s say that here we focus on the vertical component of our object’s motion, and we’ll see in a minute why we made this choice. In that case, our equation of motion would become the final velocity in the 𝑦-direction is equal to the initial velocity of our object in the 𝑦-direction plus its acceleration in the 𝑦-dimension times 𝑡 sub 𝑓.

By using this time 𝑡 sub 𝑓, we’re saying that the final velocity of our projectile is its velocity just as it comes back to Earth. As we look into its initial velocity in the 𝑦-direction, its velocity just as it leaves the potato launcher, we can indicate that using the variables in our up-close view sketch. If we let the initial trajectory of our potato be the hypotenuse of a right triangle, then the initial vertical or 𝑦-velocity of our potato is represented by this height here, and that’s equal in size to 𝑣 sub 𝑖 times the sin of the angle 𝜃.

So, 𝑣 sub 𝑖𝑦 equals 𝑣 sub 𝑖 times the sin of 𝜃, and this result can actually help us solve for 𝑣 sub 𝑓𝑦 as well. Here’s how. As our potato goes through the air, if we consider just its motion in the vertical dimension, then we can think of it as traveling up and then coming right back down. All during its flight, it’s subject to the acceleration due to gravity 𝑔. Since this is the only force affecting its motion, this tells us that the speed of our projectile just as it leaves the launcher is the same as its speed just before it lands back on Earth. On the way up, just as it goes from its initial vertical speed and decelerates to zero, so on the way down it accelerates and eventually reaches its initial vertical speed just before landing. This means that when we consider our particle’s velocity, which is a vector, we can say that the magnitude of 𝑣 sub 𝑓𝑦 is the same as that of 𝑣 sub 𝑖𝑦. But the sign is opposite. In other words, 𝑣 sub 𝑓𝑦 is negative 𝑣 sub 𝑖 times the sin of 𝜃.

And now the only thing left to fill in in this equation is a value for the acceleration of our projectile in the 𝑦-direction. But we already know that that’s 𝑔, the acceleration due to gravity. Here’s something important to keep in mind, though. Because we defined vertically up as the positive 𝑦-direction, this means our acceleration in the 𝑦-direction will have to be negative since it acts downward. So, if 𝑔 is 9.8 meters per second squared, then 𝑎 sub 𝑦 must be negative 𝑔.

Note that now we’re at a point where if we know 𝑣 sub 𝑖 and 𝜃, we can rearrange this equation to solve for 𝑡 sub 𝑓. If we first subtract 𝑣 sub 𝑖 times the sin of 𝜃 from both sides, then we get negative two 𝑣 sub 𝑖 sin 𝜃 equals negative 𝑔 times 𝑡 sub 𝑓. Dividing both sides by negative 𝑔, that factor cancels on the right and the negative signs cancel on the left. And so, we find that two times 𝑣 sub 𝑖 times the sin of 𝜃 over 𝑔 equals 𝑡 sub 𝑓, the time-in-flight of our projectile.

Let’s record this result off to the right. And now, let’s work on solving for the maximum height that our projectile achieves. Once again, this has to do with the vertical rather than the horizontal component of our object’s motion. The equation of motion we’ll choose to work with is this second one here. We choose this because it has the variable we want to solve for — remember that 𝑠 represents displacement — while we know the values for the other variables involved, in this case the vertical acceleration, the initial vertical velocity, and final vertical velocity.

Regarding that final vertical velocity of our object, we’ve called it 𝑣 sub 𝑓𝑦, we know that when our potato is at its maximum height, its position will be right here and that at that moment in time, its vertical velocity will be zero. When the potato is at its high point then, its velocity in the 𝑦-direction is zero. So, the left-hand side of this equation becomes zero. Just like before, the potato’s initial velocity in the 𝑦-direction is 𝑣 sub 𝑖 times the sin of 𝜃 and its acceleration in the 𝑦-direction is negative 𝑔. This gives us this equation. And recall it’s ℎ max we want to solve for.

What we’ll do then is subtract 𝑣 sub 𝑖 sin 𝜃 quantity squared from both sides. And from there, we divide both sides of our equation by negative two times 𝑔. This cancels that factor on the right-hand side and eliminates the negative signs on the left. So, in terms of the initial speed and direction of our projectile, the maximum height it will reach is equal to the quantity 𝑣 sub 𝑖 times the sin of 𝜃 squared all over two times 𝑔. And note that another way to write this is 𝑣 sub 𝑖 squared times the sin squared of 𝜃 over two 𝑔.

Okay, so we’ve covered time-in-flight and maximum height achieved. Now, let’s look at how to calculate the range of a projectile. We’ll do this a bit differently than before, because now the horizontal velocity of our projectile is important. Looking over at our initial zoomed-in view, we know that the initial horizontal velocity of our projectile is 𝑣 sub 𝑖 times the cos of 𝜃. And even though we’ve called this the initial horizontal velocity, this actually doesn’t change over time. Because there’s no acceleration on our projectile in the horizontal direction, its horizontal velocity is constant. This is why earlier on, we chose to use the vertical dimension in looking at this equation of motion rather than the horizontal.

In a horizontal plane, the acceleration is zero. So, we just have an equation that says 𝑣 sub 𝑖 times the cos of 𝜃 equals 𝑣 sub 𝑖 times the cos of 𝜃. That’s not so helpful. And so, that’s why we didn’t choose that dimension. But now, we do need to work with velocity in this direction because we can see it will have an impact on the range that our projectile travels. To start figuring this out, let’s choose an equation of motion that will suit our needs.

First, we want to use an equation of motion that involves the variable 𝑠; that includes three of these four. Second, it will be nice to take advantage of the fact that in the horizontal direction 𝑎 is zero; in other words, any term with 𝑎 in it will go to zero. That leaves us with this equation of motion and this one. Looking more closely, though, we can see that actually, this equation of motion isn’t an option for us. Because if 𝑎 is zero, then that cancels out the term that involves the factor we want to solve for, 𝑠. That settles it then; we’ll work with this equation of motion to solve for the range.

The equation we can write is that the range of our projectile is equal to its initial velocity in the 𝑥-direction times the total time-in-flight 𝑡 sub 𝑓 plus one-half its acceleration in the 𝑥-direction times the time-in-flight squared. As we mentioned, 𝑎 sub 𝑥 is zero because there’s no acceleration in the horizontal direction. We’ve also seen that 𝑣 sub 𝑖 in the 𝑥-direction is 𝑣 sub 𝑖 times the cos of 𝜃. And then for 𝑡 sub 𝑓, we can refer to the equation that we derived earlier. It’s two times 𝑣 sub 𝑖 times the sin of 𝜃 over 𝑔. So, the range our projectile travels is equal to this product, which is equal to two times 𝑣 sub 𝑖 squared times the cos of 𝜃 times the sin of 𝜃 all over 𝑔.

So then, we now have equations for the range, maximum height, and time-in-flight for a projectile given its initial velocity, that is, its initial speed and the direction of travel. Knowing all this, let’s get a bit of practice now through an example.

A projectile has an initial speed of 25 meters per second and is fired at an angle of 48 degrees above the horizontal. What is the time between the projectile leaving the ground and returning to the ground at the same height that it was launched from?

Okay, let’s say that this is the path that our projectile follows. We’re told that when it’s initially launched, it has a speed of 25 meters per second and that it’s fired off at an angle we’ll call 𝜃 of 48 degrees. Knowing this, we want to solve for the time it takes for this projectile to return to the surface of the Earth. In other words, how long does its full trajectory take? We’ll call this time 𝑡 sub 𝑓. And since we’re working with a projectile, that means we can recall the equation for this time.

Given a moving object affected only by the force of gravity, if that object is launched from the ground with an initial speed 𝑣 sub 𝑖 at an angle 𝜃, then its total time-in-flight is given by this expression. Since we know our object’s initial speed and its angle of launch, and we also know that 𝑔 is equal to 9.8 meters per second squared, we can plug these values into our equation for 𝑡 sub 𝑓. When we enter this expression on our calculator, to two significant figures, we find the result of 3.8 seconds. That’s the total time that this projectile is in flight.

Let’s look now at another example exercise.

A projectile is fired at an angle of 32 degrees above the horizontal with an initial speed of 44 meters per second. What is the maximum upward vertical displacement of the projectile from its launch position?

Okay, so, let’s say that this is ground level and this orange line shows us the trajectory of this projectile. The projectile is fired at an initial speed we’ll call 𝑣 sub 𝑖 and at an angle we’ll call 𝜃. We want to solve for the maximum upward vertical displacement the projectile reaches, that is, this displacement here. We’ll call it ℎ sub max.

Since we’re working with a projectile and we know its initial speed and direction of travel, we can recall an equation for the maximum height reached by a projectile in terms of its initial speed and its direction. Since we’re given 𝑣 sub 𝑖 and 𝜃 and we also know that the acceleration due to gravity is 9.8 meters per second squared, we can plug these values into our equation for ℎ sub max. When we enter this expression on our calculator, to two significant figures, we find a result of 28 meters. This is the maximum vertically upward displacement of our projectile.

Let’s look now at one last example exercise.

A projectile is fired at an angle of 66 degrees above the horizontal. The time between the projectile leaving the ground and returning to the ground at the same height that it was launched from is 2.9 seconds. What was the projectile’s Initial speed?

Okay, let’s say that this is the path that our projectile follows. The initial launch angle of this projectile, we’ll call it 𝜃, is 66 degrees. And the time it takes for our projectile to travel this full path, we’ll call this 𝑡 sub 𝑓, is 2.9 seconds. Knowing this, we want to solve for the initial speed of the projectile. We’ll call it 𝑣 sub 𝑖. Since we are working with a projectile, a body moving under the influence only of the force of gravity, we can recall that the total time it takes for a trajectory that starts and ends at the same height is equal to two times the projectile’s initial speed multiplied by the sin of its launch angle all divided by 𝑔.

In our situation, it’s not 𝑡 sub 𝑓 we want to solve for but 𝑣 sub 𝑖. To begin doing that, we can multiply both sides of this equation by 𝑔 over two times the sin of 𝜃. Over on the right-hand side, the factors of two, 𝑔, and sin 𝜃 all cancel out, leaving us with 𝑣 sub 𝑖. And we see then that 𝑣 sub 𝑖 is equal to 𝑔 times 𝑡 sub 𝑓 over two times the sin of 𝜃.

Looking at what’s given to us in our problem statement, we know 𝜃 as well as 𝑡 sub 𝑓, and we can recall further that 𝑔 equals 9.8 meters per second squared. If we then plug in these values to our equation for 𝑣 sub 𝑖, to two significant figures, we get a result of 16 meters per second. This is the initial speed our projectile would need to have at this given launch angle so that it’s in the air for 2.9 seconds.

Let’s now summarize what we’ve learned in this lesson through a few key points. We saw here that for a projectile that lands at the same height from which it was launched, if we know its initial launch speed 𝑣 sub 𝑖 and its initial launch angle 𝜃, then its total time-in-flight 𝑡 sub 𝑓 equals two times its initial speed times the sin of 𝜃 divided by 𝑔. Similarly, the maximum height above ground level that it achieves, ℎ sub max, equals its initial speed squared times the sin of this angle 𝜃 squared divided by two times 𝑔. And lastly, the projectile’s range, its total horizontal distance traveled, is equal to two times 𝑣 sub 𝑖 squared times sin 𝜃 times cos 𝜃 all over 𝑔.

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