# Video: Projectile Motion

In this video we learn what projectile motion is, how to define and set up projectile motion scenarios, and how to use to the kinematic equations to solve them.

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### 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.