Arrow's Max Height: 80 M/s At 55° - Physics Calculation

Hey guys! Ever wondered how high an arrow can go when launched at a specific speed and angle? Let's dive into the physics of projectile motion and figure out the maximum height an arrow reaches when shot at 80 m/s with an elevation angle of 55°. This is a classic physics problem that combines concepts of kinematics and trigonometry, and it's super cool once you understand the steps. We'll break it down so it's easy to follow. So, grab your thinking caps, and let's get started!

Understanding Projectile Motion

In projectile motion, understanding the initial conditions is paramount. When an arrow is launched, it doesn't just fly straight up and then fall down. It follows a curved path, known as a trajectory. This trajectory is a result of two independent motions: horizontal motion and vertical motion. The initial velocity of the arrow (80 m/s in our case) and the launch angle (55°) are crucial for determining the characteristics of this motion. The initial velocity is a vector quantity, meaning it has both magnitude (speed) and direction (angle). To analyze the motion, we need to break this initial velocity into its horizontal and vertical components. The horizontal component, ${ v_{0x} }$, remains constant throughout the flight (neglecting air resistance), while the vertical component, ${ v_{0y} }$, changes due to gravity. Gravity acts as a constant downward acceleration, slowing the arrow as it moves upward and speeding it up as it falls back down. To find the maximum height, we need to focus on the vertical motion, specifically how the vertical velocity changes over time.

The vertical component of the initial velocity (${ v_{0y} }$) is what directly opposes gravity and determines how high the arrow will go. This component is calculated using the formula ${ v_{0y} = v_0 \sin(\theta) }$, where ${ v_0 }$ is the initial velocity and ${ \theta }$ is the launch angle. In our problem, this would be ${ 80 \times \sin(55^\circ) }$. The horizontal component (${ v_{0x} }$) is calculated using ${ v_{0x} = v_0 \cos(\theta) }$, but it's not directly relevant for calculating the maximum height, as it only affects the range (horizontal distance) of the projectile. The independence of horizontal and vertical motion is a key concept here. We can analyze each component separately and then combine our findings to understand the overall motion. When the arrow reaches its maximum height, its vertical velocity momentarily becomes zero. This is the critical point we'll use in our calculations. Before we jump into the formulas, it’s important to visualize the path of the arrow. Imagine the arrow soaring upwards, gradually losing its vertical speed until it pauses at the very top before gravity pulls it back down. At that highest point, the arrow is neither moving upwards nor downwards – its vertical velocity is zero. This understanding helps make the calculations more intuitive and less like just plugging numbers into a formula.

Breaking Down Initial Velocity

The initial velocity is crucial in determining the arrow's trajectory. To effectively analyze the motion, we must decompose this initial velocity into its horizontal and vertical components. This is where trigonometry comes into play. Guys, remember SOH CAH TOA from your high school math class? It’s going to be our best friend here! The horizontal component, often denoted as ${ v_{0x} }$, represents the arrow's speed in the horizontal direction, while the vertical component, ${ v_{0y} }$, represents its initial upward speed. These components are independent of each other, meaning that gravity only affects the vertical component, and (in an idealized scenario with no air resistance) the horizontal component remains constant throughout the arrow's flight. The horizontal component ${ v_{0x} }$ is calculated using the formula: ${ v_{0x} = v_0 \cos(\theta) }$, where ${ v_0 }$ is the initial velocity (80 m/s) and ${ \theta }$ is the launch angle (55°). So, ${ v_{0x} = 80 \cos(55^\circ) }$. The vertical component ${ v_{0y} }$ is calculated using the formula: ${ v_{0y} = v_0 \sin(\theta) }$. Thus, ${ v_{0y} = 80 \sin(55^\circ) }$. Calculating these components helps us separate the complex motion into simpler, manageable parts. We can now focus on the vertical motion to determine the maximum height, as this is the component directly affected by gravity. Think of it this way: the horizontal component tells us how far the arrow will travel, while the vertical component tells us how high it will go. By isolating these components, we can apply kinematic equations more effectively. It’s like dissecting a problem into smaller pieces, making it easier to solve. Once we have these initial velocities, we can use them in our kinematic equations to find out all sorts of cool things about the arrow's flight, like how long it stays in the air and how far it travels.

Calculating Maximum Height

To calculate the maximum height, we'll use the kinematic equation that relates final velocity, initial velocity, acceleration, and displacement. This equation is particularly useful because it doesn't require us to know the time it takes to reach the maximum height. The equation is: ${ v_f^2 = v_i^2 + 2a\Delta y }$, where:

• ${ v_f }$ is the final velocity
• ${ v_i }$ is the initial velocity
• ${ a }$ is the acceleration
• ${ \Delta y }$ is the displacement (the change in vertical position)

In our case, at the maximum height, the final vertical velocity (${ v_f }$) is 0 m/s. The initial vertical velocity (${ v_i }$) is ${ v_{0y} }$, which we calculated as ${ 80 \sin(55^\circ) }$. The acceleration (${ a }$) is due to gravity, which acts downwards, so we use ${ a = -g }$, where ${ g }$ is the acceleration due to gravity (approximately 9.8 m/s²). The displacement (${ \Delta y }$) is what we're trying to find – the maximum height. Let's plug in the values: ${ 0^2 = (80 \sin(55^\circ))^2 + 2(-9.8)\Delta y }$. Now, we need to solve for ${ \Delta y }$. Rearranging the equation, we get: ${ \Delta y = -\frac{(80 \sin(55^\circ))^2}{2(-9.8)} }$. Calculating this will give us the maximum height the arrow reaches. Remember, the negative signs cancel out, giving us a positive value for height, which makes sense! This equation is powerful because it directly links the initial conditions (velocity and angle) to the outcome (maximum height). It's a testament to the elegance and predictive power of physics. Guys, it's awesome how we can use a simple equation to figure out something as cool as how high an arrow will fly, isn’t it?

Applying the Kinematic Equation

Okay, let's get into the nitty-gritty of applying the kinematic equation. We've already established the equation we'll use: ${ v_f^2 = v_i^2 + 2a\Delta y }$. The key here is to correctly identify each term and plug in the values with the correct signs. Remember, physics is all about the details! We know that at the maximum height, the final vertical velocity (${ v_f }$) is 0 m/s. This is because the arrow momentarily stops moving upwards before it starts falling back down. The initial vertical velocity (${ v_i }$) is ${ 80 \sin(55^\circ) }$, which we need to calculate. Using a calculator, ${ \sin(55^\circ) }$ is approximately 0.819. So, ${ v_i = 80 \times 0.819 \approx 65.52 \text{ m/s} }$. The acceleration (${ a }$) is the acceleration due to gravity, which is approximately -9.8 m/s². We use a negative sign because gravity acts downwards, opposing the upward motion of the arrow. The displacement (${ \Delta y }$) is the maximum height we're trying to find. Now, let's plug these values into the equation: ${ 0^2 = (65.52)^2 + 2(-9.8)\Delta y }$. Simplifying, we get: ${ 0 = 4292.87 - 19.6\Delta y }$. To solve for ${ \Delta y }$, we rearrange the equation: ${ 19.6\Delta y = 4292.87 }$. Dividing both sides by 19.6, we get: ${ \Delta y = \frac{4292.87}{19.6} \approx 219.02 \text{ meters} }$. So, the maximum height the arrow reaches is approximately 219.02 meters. Wow, that's pretty high! This calculation demonstrates the power of physics in predicting real-world outcomes. By understanding the principles of projectile motion and using kinematic equations, we can accurately determine the trajectory and key parameters of a projectile's flight. It's like having a superpower – we can predict the future of the arrow! It also shows how important it is to keep track of units and signs in physics calculations. A small mistake can lead to a big difference in the final answer. So, always double-check your work, guys!

Final Answer and Conclusion

Alright, let's wrap things up! The final answer for the maximum height the arrow reaches is approximately 219.02 meters. That's seriously impressive! We've journeyed through the concepts of projectile motion, broken down the initial velocity into its components, and applied a kinematic equation to arrive at this result. Physics, guys, is awesome! We've seen how a combination of trigonometry and kinematics can help us understand and predict the motion of objects in the real world. This problem is a perfect example of how theoretical physics can be applied to practical scenarios. Whether it's calculating the trajectory of an arrow, a baseball, or even a rocket, the fundamental principles remain the same. The key is to understand the underlying physics, break the problem down into manageable steps, and apply the appropriate equations. And remember, practice makes perfect! The more problems you solve, the more comfortable you'll become with these concepts. So, keep exploring, keep questioning, and keep calculating! Who knows? Maybe you'll be the one designing the next generation of archery equipment or even spacecraft. The possibilities are endless when you have a solid understanding of physics. And remember, understanding how things move is not just about solving problems in a textbook; it's about understanding the world around us. So, next time you see an arrow flying through the air, you'll have a whole new appreciation for the physics at play.