Solving Projectile Motion: Finding The Ball's Horizontal Distance

Hey guys! Let's dive into a fun math problem involving a ball being thrown! We're going to use a quadratic equation to figure out how far the ball travels horizontally before it hits the ground. This is a classic example of projectile motion, and it's super cool how math helps us understand the real world. So, imagine this: you throw a ball, and it follows a curved path. We want to figure out how far that ball goes before it lands. We'll use a special equation to represent this path. This type of equation is called a quadratic equation, and its graph is a U-shaped curve, or parabola. In our case, the ball's height, $f(x)$, is modeled by the equation $f(x) = -0.2x^2 + 1.4x + 6$. Here, x represents the horizontal distance the ball travels, and $f(x)$ represents the height of the ball at that horizontal distance. This means that the equation describes the ball's trajectory. The ball starts at a height of 6 feet, as we see from the equation. The negative sign in front of the $x^2$ tells us that the parabola opens downwards, which is what we expect since gravity pulls the ball back down. Our main goal is to find the horizontal distance x when the ball hits the ground. When the ball hits the ground, its height $f(x)$ is zero. So, we need to solve the equation for when $f(x) = 0$. This brings us to the core of our problem: solving the quadratic equation. It's a fundamental concept in algebra, and it is important to understand projectile motion. Once we understand that concept, we can utilize it to solve a ton of physics-related problems. Now, let's break down how we can solve this. Are you ready to solve some equations? Let's get to it!

Setting Up the Equation and Understanding the Components

Okay, so the problem says we have this ball thrown, and its path is defined by the equation $f(x) = -0.2x^2 + 1.4x + 6$. Remember, $x$ is the horizontal distance, and $f(x)$ is the height. To find out when the ball hits the ground, we need to know the horizontal distance when the height is zero. This means we want to find the value(s) of x when $f(x) = 0$. So, we set up the equation like this: $0 = -0.2x^2 + 1.4x + 6$. This is our starting point. Now we've got a quadratic equation in the standard form $ax^2 + bx + c = 0$, where $a = -0.2$, $b = 1.4$, and $c = 6$. The coefficient a is negative, which means the parabola opens downwards. The coefficient b influences the position and the slope of the parabola. The coefficient c is the point where the parabola intersects with the y-axis. We have a few different ways we can solve this quadratic equation. We could try factoring, but in this case, it might not be the easiest method. Alternatively, we could use the quadratic formula. The quadratic formula is a powerful tool that can solve any quadratic equation! The formula is: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. It's a formula that always works, no matter what values a, b, and c have. It is designed to directly give us the roots of the quadratic equation, which are the values of x where the parabola crosses the x-axis. These are the points where the height is zero. Now, let's substitute the values of $a$, $b$, and $c$ from our original equation into this formula. This is a classic algebra problem. You will encounter this again and again in your math journey. So, it's super important to master it.

Applying the Quadratic Formula

Alright, let's plug our values into the quadratic formula: $x = \frac{-1.4 \pm \sqrt{1.4^2 - 4(-0.2)(6)}}{2(-0.2)}$. Now, let's carefully calculate the parts inside the formula step by step. First, calculate $1.4^2$, which is $1.96$. Next, let's figure out $4(-0.2)(6)$, which is equal to $-4.8$. The discriminant, $b^2 - 4ac$, becomes $1.96 - (-4.8)$, which simplifies to $1.96 + 4.8 = 6.76$. Then, we take the square root of 6.76, which is 2.6. Now we have $x = \frac{-1.4 \pm 2.6}{-0.4}$. Remember that the $\pm$ symbol means we're going to have two possible solutions – one where we add 2.6 and one where we subtract 2.6. Let's do both calculations. First, we'll add: $x = \frac{-1.4 + 2.6}{-0.4} = \frac{1.2}{-0.4} = -3$. Next, we'll subtract: $x = \frac{-1.4 - 2.6}{-0.4} = \frac{-4}{-0.4} = 10$. So, we have two possible values for x: -3 and 10. Since x represents the horizontal distance, a negative value doesn't make sense in this real-world context. It's like saying the ball traveled a negative distance. That's why we can ignore the negative answer. So, that means the ball travels a horizontal distance of 10 feet before hitting the ground. Remember, math always gives us solutions, and it's up to us to interpret them in the context of the problem. This helps us understand when our solutions make sense and when they don't.

Interpreting the Results and Considering Real-World Context

Okay, we crunched the numbers, and we got two answers: x = -3 and x = 10. Now comes the part where we need to think about what these answers actually mean in the real world. Remember, x represents the horizontal distance the ball travels from where it was thrown. So, a negative distance doesn't make a lot of sense here. You can't really throw a ball a negative distance, right? It would be like going backward in time! This means the x = -3 solution is not realistic in our scenario. It's a mathematical solution to the equation, but it doesn't fit the physical situation we're describing. This leaves us with x = 10 feet. This solution makes perfect sense. It tells us that the ball travels 10 feet horizontally before it hits the ground. This is the answer we're looking for! The ball's trajectory starts at a height of 6 feet. It goes up, curves, and then comes back down. The 10 feet represents the total horizontal distance the ball covers from the point it was thrown to the point it lands. In real-world projectile motion, there are a lot of things that can affect the ball's path, such as air resistance and wind. But in this simplified mathematical model, we're ignoring those factors. The model gives us a pretty good approximation, but it's not perfect. The key takeaway is that we used the quadratic equation to model the ball's path and then used the quadratic formula to solve for when the ball hits the ground. This is a valuable skill in many areas of math and science. Remember to always consider the real-world implications of your solutions, and to make sure they make sense in the context of the problem.

Conclusion

So, there you have it, guys! We've successfully solved the problem and found the horizontal distance the ball travels. We used a quadratic equation to model the ball's path and then used the quadratic formula to find the solution. We found that the ball travels 10 feet horizontally before hitting the ground. This shows us how powerful math can be in modeling real-world scenarios. Keep practicing, and you'll get better and better at solving these types of problems. I hope you enjoyed this explanation, and it was helpful. Feel free to ask questions or try out more problems on your own. Keep practicing and you will get better! Good luck!