### Video Transcript

A dart is thrown horizontally at a speed of 10 meters per second at the bull’s-eye of a dartboard, 2.4 meters away. How far below the intended target does the dart hit?

Let’s draw a picture of what’s going on in the situation.

We have a dart being thrown at a dartboard that’s 2.4 meters away, and the initial speed of the dart is 10 meters per second. Now the dart is thrown so that if it followed the dotted line we’ve shown, it would go and hit the bull’s-eye of the dartboard. But the real path a dart follows looks like this, where the dart actually hits the wall some distance below the bull’s-eye. We’ll call that distance 𝑑.

𝑑 is exactly what we’re trying to solve for in this problem. So let’s go ahead and solve for that distance. To begin, let’s recall a fact about projectile motion, and that fact is that motion in the horizontal direction and motion in the vertical direction are independent of one another.

What this means is we can focus entirely on motion in the horizontal direction without having to take into account the fact that the dart is dropping as it travels. Focusing on the horizontal motion will let us figure out how much time it takes from the moment the dart is released to when it hits the wall.

To figure that out, let’s recall an equation that we’ve seen earlier. That equation states that the speed of an object, represented by 𝑠, is equal to the distance it travels divided by the time it takes to travel that distance. We can use this fact of motion independence to focus entirely on the horizontal direction, which will let us solve for the length of time that the dart is in the air. To begin making progress in this direction, let’s recall that the speed an object travels is equal to the distance that the object travels divided by the time it takes to move that distance.

When we apply this relationship to our own scenario, we see that 𝑣 sub 𝑥 is what we would use in place of 𝑠. And in place of 𝑑, we would use our given distance of 2.4 meters. And all in all, this will let us solve for the time that the dart is in the air. When we rearrange this relationship to solve for time, we see that time is equal to the distance the object travels, in this case our dart travelling a distance of 2.4 meters, divided by the speed at which it travels that distance, which we’re given as 10 meters per second.

We can see from this numerator and denominator that overall the time the dart is in the air is equal to 0.24 seconds. Now that we’ve figured that out, let’s switch over from motion in the 𝑥 or horizontal direction to motion in the vertical direction and consider what our kinematic equations are for motion in a field affected by gravity.

Those kinematic equations you may recall are these four. Now take a look at these four equations and keep an eye out for equations that match information we have, either by solving for it or being given it, and also what we’re trying to solve for, which recall is the vertical distance that this dart drops over the course of its flight.

Looking over these kinematic equations, the third one down may stand out to you. It solves for 𝑑, a distance, and that’s what we wanna solve for. And in this first term, we see that because our initial velocity in the 𝑦 Direction is zero in our example, the first term cancels out. Specifically, it goes to zero because that initial velocity itself is zero. And in our second term, we have a 𝑡 squared value, and we just solved for time so we’re able to figure out what that is.

So let’s rewrite this third kinematic equation in terms of variables for our problem. And what we see is that the distance the dart drops is equal to one-half times 𝑔, 9.8 meters per second squared, times our time squared, 0.24 seconds.

So when we calculate all this together, we find a total vertical drop of our dart of 0.28 meters. That is how far below the bull’s-eye our dart would actually run into the wall.