Firework Maximum Height Calculation A Step-by-Step Physics Guide

Hey guys! Today, we're diving into a classic physics problem that combines the principles of kinematics and projectile motion. We're going to figure out how to calculate the maximum height a firework reaches when it's launched straight up into the air. This problem is super relevant because it helps us understand how objects move under the influence of gravity, which is a fundamental concept in physics. Let's break down the problem step by step, so you can master this type of calculation. We'll start with understanding the core concepts and then move on to the actual math involved. By the end, you’ll be able to tackle similar problems with confidence. So, let’s get started and make sure we cover all the bases to make this crystal clear for everyone!

Understanding the Physics Behind the Firework

Before we jump into the math, let's make sure we all understand the physics concepts at play here. The main idea is that the firework is launched upward with an initial velocity, but gravity is constantly pulling it back down. This constant pull of gravity causes the firework to slow down as it rises until it momentarily stops at its highest point. Then, it starts falling back to the ground, accelerating due to gravity. It's like throwing a ball straight up – it goes up, slows down, stops, and then comes back down. In our problem, we're given a few key pieces of information. First, the firework's initial velocity is 128 feet per second (ft/s). This is how fast it's moving upward when it leaves the ground. Second, we know that the acceleration due to gravity is 32 feet per second squared (ft/s²). This means that every second, the firework's upward velocity decreases by 32 ft/s. Gravity is the force that slows down the firework as it ascends. Understanding these concepts is crucial because they form the foundation for solving the problem. We need to consider how gravity affects the firework’s motion, causing it to decelerate until it reaches its peak height. At that point, its velocity will be zero before it starts to fall back down. Knowing the initial velocity and the acceleration due to gravity allows us to predict the firework’s behavior and, most importantly, calculate its maximum height. So, with these basics in mind, let's move on to the equation that will help us put these concepts into practice. Let's keep breaking it down together!

The Equation We'll Use

The equation we'll use to solve this problem is a fundamental one in physics, specifically in kinematics, which deals with the motion of objects. The equation is:

$$s = at^2 + vt + h_0$$

Where:

• s is the displacement (the final position) of the object
• a is the acceleration
• t is the time
• v is the initial velocity
• h₀ is the initial height

This equation is a second-degree polynomial, and it perfectly describes the motion of an object under constant acceleration, such as a firework under the influence of gravity. Now, let’s break down each part of this equation in the context of our problem. First, s represents the final height of the firework at any given time t. In our case, we are interested in the maximum height, which will be the highest value of s. Next, a stands for acceleration, which, in our scenario, is the acceleration due to gravity. Since gravity is pulling the firework downwards, we'll use a negative value for a, specifically -32 ft/s². The term t represents the time elapsed since the firework was launched. We'll need to find the time at which the firework reaches its maximum height. The initial velocity v is the speed at which the firework was launched upwards, which is given as 128 ft/s. This is the initial "push" that gets the firework moving against gravity. Finally, h₀ is the initial height of the firework. Since the firework is launched from ground level, h₀ is 0. Understanding each component of this equation is crucial because it allows us to plug in the known values and solve for the unknowns, in this case, the maximum height s. By rearranging and applying this equation correctly, we can determine exactly how high the firework will go before it starts to fall back down. So, let’s dive deeper into how we can use this equation to solve our specific problem. Are you guys ready to get into the calculations?

Applying the Equation to Our Problem

Okay, let's put our knowledge into action and apply the equation to our firework problem. We have the equation:

$$s = at^2 + vt + h_0$$

We know the following values:

• a (acceleration due to gravity) = -32 ft/s² (negative because it acts downwards)
• v (initial velocity) = 128 ft/s
• h₀ (initial height) = 0 ft (ground level)

So, we can rewrite the equation with these values plugged in:

$$s = -32t^2 + 128t + 0$$

This simplifies to:

$$s = -32t^2 + 128t$$

Now, our goal is to find the maximum height s. This occurs at the vertex of the parabola described by the equation. In other words, we need to find the time t at which the firework reaches its peak. There are a couple of ways to do this. One way is to use calculus, but we'll stick to algebra here to keep things simple. The time at which the maximum height is reached can be found using the formula for the vertex of a parabola, which is t = -b / 2a, where a and b are the coefficients in the quadratic equation. In our equation, a = -32 and b = 128. So, let's plug these values in:

$$t = -128 / (2 * -32)$$

$$t = -128 / -64$$

$$t = 2$$

So, the firework reaches its maximum height at t = 2 seconds. Now that we know the time, we can plug this value back into our equation to find the maximum height s:

$$s = -32(2)^2 + 128(2)$$

$$s = -32(4) + 256$$

$$s = -128 + 256$$

$$s = 128$$

Therefore, the maximum height reached by the firework is 128 feet. See how we broke it down step by step? First, we identified the known values, then we plugged them into the equation, and finally, we solved for the unknown. This systematic approach is key to tackling physics problems. Now, let's summarize our findings and discuss the significance of this result.

Calculating the Maximum Height

Let's recap the calculation we just went through to make sure everything is crystal clear. We started with the equation:

$$s = -32t^2 + 128t$$

We determined that the time at which the firework reaches its maximum height is t = 2 seconds by using the vertex formula t = -b / 2a. Now, let's plug that time back into the equation to find the maximum height:

$$s = -32(2)^2 + 128(2)$$

First, we calculate 2^2, which is 4:

$$s = -32(4) + 128(2)$$

Next, we multiply -32 by 4:

$$s = -128 + 128(2)$$

Then, we multiply 128 by 2:

$$s = -128 + 256$$

Finally, we add -128 and 256:

$$s = 128$$

So, the maximum height s is 128 feet. This means the firework travels 128 feet upwards before it starts to fall back down. Breaking down the calculation like this helps to ensure we follow each step correctly and understand the process. It also makes it easier to spot any potential mistakes. Each step is logical and builds on the previous one, leading us to the final answer. Remember, it's important to be methodical when solving physics problems. Now that we've clearly shown the calculation, let's discuss what this result means in the context of the problem. Understanding the significance of the answer is just as important as getting the correct number. It helps us connect the math to the real-world scenario and reinforces our understanding of the underlying physics principles.

Conclusion: What Does This All Mean?

So, we've crunched the numbers and found that the firework reaches a maximum height of 128 feet. That's pretty high! But what does this mean in a broader context? Well, this problem illustrates the fundamental principles of projectile motion. We've seen how gravity affects an object launched upwards, slowing it down until it reaches its peak, and then pulling it back down. This is a common scenario in physics, whether we're talking about fireworks, balls thrown in the air, or even rockets launched into space. The key takeaways here are the concepts of initial velocity, acceleration due to gravity, and how these factors combine to determine the trajectory of an object. The equation we used, s = at^2 + vt + h_0, is a powerful tool for predicting the motion of objects under constant acceleration. By understanding this equation and how to apply it, you can solve a wide range of physics problems. Another important aspect of this problem is the methodical approach we took to solve it. We identified the known values, plugged them into the equation, and then carefully worked through the calculations step by step. This systematic approach is crucial for success in physics and other problem-solving disciplines. It helps to avoid errors and ensures that we understand each step of the process. Guys, I hope this detailed explanation has helped you understand how to calculate the maximum height of a firework (or any projectile) launched into the air. Remember, the key is to break the problem down into smaller steps, understand the underlying physics principles, and apply the appropriate equations. With practice, you'll become more confident in your ability to tackle these types of problems. Keep practicing and exploring, and you'll be amazed at what you can achieve!