Max Height Of Projectile Launched At 32 Degrees

Hey guys! Let's dive into a classic physics problem involving projectile motion. We're going to break down how to calculate the maximum height reached by a projectile when we know its initial velocity and launch angle. This is a fundamental concept in physics, and understanding it will help you analyze all sorts of real-world scenarios, from a baseball soaring through the air to a rocket blasting off into space.

Understanding Projectile Motion

Before we jump into the calculations, let's make sure we're all on the same page about projectile motion. Simply put, it's the motion of an object thrown or projected into the air, subject only to the acceleration of gravity. We're neglecting air resistance here to keep things simple. The path the projectile follows is a parabola, which means it curves upwards and then back down due to gravity's pull. Key factors affecting projectile motion include the initial velocity, the launch angle, and of course, gravity. The initial velocity is how fast the object is launched, and the launch angle is the angle at which it's launched relative to the horizontal. Gravity, which always acts downwards, constantly decelerates the projectile's vertical motion.

Breaking Down Initial Velocity

The initial velocity is crucial in determining the projectile's trajectory. It has both horizontal and vertical components. The horizontal component (Vx) is what carries the projectile forward, and the vertical component (Vy) is what propels it upwards against gravity. We can find these components using trigonometry:

• Vx = V * cos(θ)
• Vy = V * sin(θ)

Where V is the initial velocity and θ is the launch angle. The horizontal component remains constant throughout the projectile's flight (neglecting air resistance), while the vertical component changes due to gravity. As the projectile moves upwards, gravity slows it down until it momentarily stops at its maximum height. Then, gravity pulls it back down, increasing its vertical speed.

The Role of Gravity

Gravity is the unsung hero (or villain, depending on your perspective!) of projectile motion. It's the force that causes the projectile to curve downwards and eventually return to the ground. The acceleration due to gravity is approximately 9.8 m/s² (often rounded to 10 m/s² for simplicity) and it acts only in the vertical direction. This means that the horizontal velocity of the projectile remains constant, but its vertical velocity changes constantly. As the projectile rises, gravity slows it down, decreasing its upward velocity. At the peak of its trajectory, the vertical velocity is momentarily zero. As the projectile falls, gravity accelerates it downwards, increasing its downward velocity. This constant interplay between initial vertical velocity and gravitational acceleration is what dictates the maximum height the projectile will reach.

Problem Setup: Launch Velocity and Angle

Alright, let's get back to the specific problem. We have a projectile launched with an initial velocity of 35 m/s at a launch angle of 32 degrees. Our mission, should we choose to accept it, is to find the maximum height the projectile reaches. This means we need to figure out the highest point in its parabolic trajectory. We'll be using the principles of projectile motion we just discussed to crack this code.

Identifying Key Information

First, let's extract the key information from the problem statement. We know:

• Initial velocity (V): 35 m/s
• Launch angle (θ): 32 degrees

We also know that the acceleration due to gravity (g) is approximately 9.8 m/s². Remember, gravity acts downwards, so we'll consider it negative in our calculations when dealing with upward motion. The key to finding the maximum height is understanding what happens at that specific point in the projectile's flight. At the maximum height, the vertical velocity (Vy) of the projectile is momentarily zero. This is because the projectile has slowed down due to gravity to a complete stop in the vertical direction before it starts falling back down.

Strategic Approach

Now that we have all the pieces of the puzzle, let's map out our strategy. We need to find the maximum height (H). We know the initial velocity, launch angle, and acceleration due to gravity. We also know that the final vertical velocity at the maximum height is zero. We can use kinematic equations, which are equations that describe motion with constant acceleration, to solve this problem. Specifically, we'll use the following kinematic equation:

vf² = vi² + 2 * a * Δy

Where:

• vf is the final velocity
• vi is the initial velocity
• a is the acceleration
• Δy is the displacement (in our case, the maximum height H)

This equation is perfect for our situation because it relates final velocity, initial velocity, acceleration, and displacement, all of which we either know or are trying to find.

Calculating Maximum Height: Step-by-Step

Alright, let's get our hands dirty with the calculations! We're going to plug in the values we have into the kinematic equation and solve for the maximum height. This is where the magic happens, so pay close attention!

Step 1: Find the Initial Vertical Velocity (Vy)

Remember, we need the initial vertical velocity (Vy) because that's the component of velocity working against gravity to propel the projectile upwards. We can calculate Vy using the following formula:

Vy = V * sin(θ)

Where:

• V = 35 m/s (initial velocity)
• θ = 32 degrees (launch angle)

Plugging in the values, we get:

Vy = 35 m/s * sin(32°)

Vy ≈ 35 m/s * 0.53

Vy ≈ 18.55 m/s

So, the initial vertical velocity of the projectile is approximately 18.55 m/s. This means it's launched upwards with a speed of 18.55 meters per second.

Step 2: Apply the Kinematic Equation

Now we're ready to use the kinematic equation:

vf² = vi² + 2 * a * Δy

We know:

• vf = 0 m/s (final vertical velocity at maximum height)
• vi = 18.55 m/s (initial vertical velocity)
• a = -9.8 m/s² (acceleration due to gravity, negative because it acts downwards)
• Δy = H (maximum height, which we're trying to find)

Plugging these values into the equation, we get:

0² = (18.55 m/s)² + 2 * (-9.8 m/s²) * H

Step 3: Solve for H (Maximum Height)

Let's simplify the equation and solve for H:

0 = 344.1025 m²/s² - 19.6 m/s² * H

Now, isolate H:

19.6 m/s² * H = 344.1025 m²/s²

H = 344.1025 m²/s² / 19.6 m/s²

H ≈ 17.56 m

Therefore, the maximum height reached by the projectile is approximately 17.56 meters. Awesome, we did it!

Result: Maximum Height Reached

Boom! We've successfully calculated the maximum height reached by the projectile. The final answer is approximately 17.56 meters. This means the projectile soared almost 18 meters into the air before gravity brought it back down. This calculation demonstrates the power of understanding projectile motion principles and applying the correct kinematic equations.

Interpretation of the Result

It's important to understand what this result tells us. The maximum height of 17.56 meters represents the peak of the projectile's trajectory. At this point, the projectile's vertical velocity is momentarily zero before it starts accelerating downwards due to gravity. The higher the initial vertical velocity and the smaller the effect of gravity (relatively speaking), the higher the projectile will go. Factors like air resistance, which we've ignored in this simplified model, would reduce the maximum height in a real-world scenario.

Practical Applications

Understanding projectile motion is incredibly useful in many real-world applications. Think about sports like baseball, basketball, or soccer, where understanding the trajectory of a ball is crucial. Engineers also use these principles to design things like cannons, rockets, and even water fountains! The concepts we've discussed here form the foundation for analyzing complex motion scenarios and making accurate predictions. This isn't just about solving physics problems; it's about understanding how the world around us works.

Conclusion: Mastering Projectile Motion

So, there you have it! We've successfully calculated the maximum height of a projectile using its initial velocity, launch angle, and the principles of projectile motion. By breaking down the problem into smaller steps, understanding the role of gravity, and applying the appropriate kinematic equations, we were able to arrive at the solution. This is a testament to the power of physics in explaining and predicting the world around us. Keep practicing these concepts, guys, and you'll become masters of projectile motion in no time!

Remember, the key takeaways from this problem are:

• Projectile motion involves an object moving under the influence of gravity.
• The initial velocity has horizontal and vertical components.
• Gravity acts only in the vertical direction.
• At maximum height, the vertical velocity is zero.
• Kinematic equations can be used to solve for unknowns in projectile motion problems.

Keep exploring, keep learning, and keep applying these principles to the world around you. You'll be amazed at what you can understand and predict!