Projectile Motion: Calculating Time To Reach A Specific Height

Hey there, math enthusiasts! Today, we're diving into the exciting world of projectile motion. We'll be tackling a classic problem: calculating the time it takes for a projectile launched upwards to reach a specific height. This involves understanding the principles of physics, particularly how gravity affects an object's vertical movement. We will use the formula that explains the height of the object based on time. Let's break it down in a way that's easy to grasp.

Understanding the Problem: The Upward Journey

Imagine you launch a ball straight up into the air. It soars upwards, its speed gradually decreasing until it momentarily stops at its highest point. Then, gravity takes over, pulling it back down. Our task is to figure out exactly when, during its ascent, the ball will reach a height of 200 feet. In this specific scenario, we're given the initial velocity and the acceleration due to gravity. The initial velocity represents how fast the ball is moving upwards when it's launched, and the acceleration due to gravity is the constant pull downwards that slows the ball's ascent. The problem provides us with the formula for the height of the ball, which we'll use to solve the problem. Essentially, we are working with a quadratic equation, where time is the unknown variable, and we are solving for time in seconds.

Now, let's look at the given parameters: the projectile is launched from the ground with an initial velocity of 120 ft/s. The acceleration due to gravity is 6 ft/s². And we need to find the time t when the height h(t) reaches 200 ft. This is a classic application of kinematics, the branch of physics that deals with the motion of objects without considering the forces that cause the motion. Kinematics provides the mathematical framework to describe position, velocity, and acceleration as functions of time. This requires an understanding of the relationship between displacement, velocity, acceleration, and time. Furthermore, understanding the concept of constant acceleration due to gravity is also crucial. This ensures that the equations of motion can be applied, as the acceleration remains constant throughout the object's flight. Finally, using quadratic equations to solve this problem is important. Because the formula that describes the height of the object is a quadratic function, solving this will help us determine the time when the height of the object is 200 ft.

The Formula: Unveiling the Height Equation

The fundamental equation that governs this type of motion is: h(t) = -at² + vt, where h(t) is the height at time t, a is the acceleration due to gravity, and v is the initial velocity. We can use this general form to model the motion of a projectile. The quadratic nature of the equation reflects the parabolic trajectory of the object under constant acceleration due to gravity. So, the height equation describes the position of the object at any given time. This height equation is critical for determining when the object will reach a particular altitude. The negative sign in front of the at² term is important, as it indicates that the acceleration due to gravity is acting downwards, thus opposing the initial upward velocity. The vt term represents the effect of the initial velocity on the height. To solve the problem, we must understand how to apply the given values to solve for the unknown variable, time, in order to get the desired solution. Remember, the formula is specifically designed to describe the vertical motion of the projectile and does not account for any horizontal movement or other external forces like air resistance. Therefore, the formula is a simplification of real-world conditions, providing an idealized view of projectile motion.

In our case, the provided formula for height is h(t) = -6t² + 120t. Notice how the acceleration due to gravity (6 ft/s²) is multiplied by t², and the initial velocity (120 ft/s) is multiplied by t.

Setting Up the Equation: Reaching the Target Height

Our goal is to find the time t when the height h(t) equals 200 ft. So, we'll substitute 200 for h(t) in our formula, transforming it into the equation 200 = -6t² + 120t. Now, we want to rearrange this equation into a standard quadratic form, which is at² + bt + c = 0. To do this, we subtract 200 from both sides: 0 = -6t² + 120t - 200. We can simplify this equation by dividing every term by -2, which results in 0 = 3t² - 60t + 100. This simplification process is helpful, as it makes the equation easier to solve. The next step is to use the quadratic formula to solve for t. The quadratic formula is a reliable method for finding the roots of a quadratic equation. It's a fundamental tool in mathematics that allows you to calculate the values of x that satisfy the equation. Using the quadratic formula guarantees that we will determine the accurate values of time when the projectile reaches a height of 200 ft. Remember to clearly identify the coefficients a, b, and c from the simplified equation before applying the formula. This is a common and important mathematical technique used in many scientific and engineering applications, showcasing how theoretical concepts can be applied to real-world problems. Keep in mind that depending on the scenario, the quadratic equation can have two solutions, one solution, or no real solutions.

Solving for Time: Using the Quadratic Formula

The quadratic formula is a lifesaver for solving equations of this type. It's given by: t = (-b ± √(b² - 4ac)) / 2a. In our simplified equation, 3t² - 60t + 100 = 0, we have a = 3, b = -60, and c = 100. Let’s plug these values into the formula.

t = (60 ± √((-60)² - 4 * 3 * 100)) / (2 * 3) t = (60 ± √(3600 - 1200)) / 6 t = (60 ± √2400) / 6

Calculating the square root of 2400, we get approximately 48.99. Therefore,

t = (60 + 48.99) / 6 or t = (60 - 48.99) / 6

Solving further, we have

t ≈ 18.16 seconds or t ≈ 1.83 seconds

These two values represent the two times when the object reaches the height of 200 ft. The smaller time (approximately 1.83 seconds) is when the object is on its way up, and the larger time (approximately 18.16 seconds) is when the object is coming back down. Therefore, the object will reach a height of 200 ft at approximately 1.83 seconds on its way up and again at approximately 18.16 seconds on its way down.

Interpreting the Results: Two Points in Time

Notice that we have two solutions for time. This is because the projectile reaches the height of 200 feet twice: once on its way up and once on its way down. The first time (1.83 seconds) is when the object is ascending, and the second time (18.16 seconds) is when it's descending. Understanding these two points is crucial for fully grasping the trajectory of the projectile. The trajectory of a projectile is a parabola, which has a symmetrical shape. The vertex of the parabola is the point where the projectile reaches its maximum height. The projectile's time to reach its maximum height can be found by using the initial velocity and acceleration due to gravity, and it is a valuable measure in understanding projectile motion. The symmetric nature of the parabola means that the time it takes to go up to a certain height is the same as the time it takes to come down to that same height. Therefore, the two solutions we found reflect this symmetry. This also underscores the significance of the quadratic equation in modelling real-world physical scenarios. Furthermore, the solutions underscore the importance of understanding the physical context when interpreting mathematical results.

Conclusion: Mastering Projectile Motion

So, there you have it! We've successfully calculated the time it takes for our projectile to reach a height of 200 feet, showcasing how to apply physics principles and mathematical formulas to solve a real-world problem. By understanding the concepts of initial velocity, acceleration due to gravity, and the quadratic equation, we can analyze and predict the motion of objects in flight. Remember, the journey of a projectile is governed by gravity, and this understanding can be applied to different scenarios and variations. This exercise underscores the interconnectedness of mathematics and physics, as you can see how mathematical models can be used to describe and predict physical phenomena. Keep practicing, and you'll find yourself confidently navigating the exciting world of projectile motion!

For more in-depth explanations and examples, check out these resources:

• Khan Academy: A great platform for understanding the basics of projectile motion and related concepts. Khan Academy