Projectile Motion Analysis How To Find Time To Maximum Height For H(t)=-16t^2+96t+6

In the realm of physics and mathematics, understanding the trajectory of objects launched into the air is a fundamental concept. This article delves into the function h(t) = -16t^2 + 96t + 6, which represents the height of an object projected from a cannon. We will explore how to determine the maximum height reached by the object and, more importantly, how long it takes to reach that peak. This involves analyzing the quadratic function and applying concepts from algebra and calculus to solve the problem.

The function h(t) = -16t^2 + 96t + 6 is a quadratic equation, where h(t) represents the height of the object at time t. The coefficients of this equation provide valuable information about the projectile's motion. The leading coefficient, -16, is related to the acceleration due to gravity, indicating that the object's trajectory is a parabola opening downwards. This downward curve signifies that the object will reach a maximum height before descending back to the ground. The term 96t represents the initial upward velocity of the object, and the constant term 6 represents the initial height from which the object was launched.

To fully grasp the behavior of this function, let's break down each component: -16t^2: This term accounts for the effect of gravity on the object. The negative sign indicates that gravity is pulling the object downwards, reducing its height over time. The coefficient 16 is half the acceleration due to gravity (approximately 32 feet per second squared), a standard value used in physics for objects near the Earth's surface. 96t: This term represents the initial upward velocity of the object. The coefficient 96 indicates the object's initial speed in feet per second. As time progresses, this term contributes positively to the height, but its effect diminishes as gravity pulls the object down. +6: This constant term represents the initial height of the object when it was launched. In this case, the object started at a height of 6 feet. Understanding these components is crucial for analyzing the projectile's motion. The quadratic nature of the function means that the trajectory will be a parabola, and the maximum height corresponds to the vertex of this parabola. Finding the vertex will give us both the maximum height and the time at which it is reached.

The problem states that the maximum height reached by the object is 150 feet. This information serves as a crucial benchmark for our calculations and understanding of the function's behavior. In mathematical terms, the maximum height corresponds to the vertex of the parabola represented by the quadratic function h(t) = -16t^2 + 96t + 6. The vertex is the highest point on the parabola, and its y-coordinate (in this case, h(t)) represents the maximum height. Knowing that the maximum height is 150 feet allows us to verify our calculations and ensure that our approach is correct. It also gives us a concrete value to compare against when we find the vertex using algebraic methods. There are several methods to find the maximum height and the time at which it occurs. One common approach is to complete the square, which transforms the quadratic equation into vertex form, making it easy to identify the coordinates of the vertex. Another method involves using calculus, specifically finding the critical points of the function by taking the derivative and setting it equal to zero. Both methods will lead to the same result, providing a robust way to confirm our findings. By understanding that the maximum height is 150 feet, we have a clear target to aim for in our calculations. This helps us stay on track and ensures that our final answer is both accurate and meaningful within the context of the problem.

To find the time at which the object reaches its maximum height, we can employ several methods, each leveraging different mathematical principles. These methods include completing the square, using the vertex formula, and applying calculus. Understanding these approaches not only helps in solving this specific problem but also enhances our broader mathematical toolkit.

Completing the Square

Completing the square is an algebraic technique used to rewrite a quadratic equation in vertex form, which is h(t) = a(t - h)^2 + k, where (h, k) represents the vertex of the parabola. The vertex form directly reveals the maximum or minimum value of the quadratic function and the time at which it occurs. To complete the square for h(t) = -16t^2 + 96t + 6, we follow these steps: Factor out the coefficient of the t^2 term from the first two terms: h(t) = -16(t^2 - 6t) + 6. Complete the square inside the parenthesis: To do this, take half of the coefficient of the t term (-6), square it ((-3)^2 = 9), and add and subtract it inside the parenthesis: h(t) = -16(t^2 - 6t + 9 - 9) + 6. Rewrite the expression inside the parenthesis as a perfect square: h(t) = -16((t - 3)^2 - 9) + 6. Distribute the -16: h(t) = -16(t - 3)^2 + 144 + 6. Simplify: h(t) = -16(t - 3)^2 + 150. From this vertex form, h(t) = -16(t - 3)^2 + 150, we can see that the vertex of the parabola is (3, 150). This means the maximum height of 150 feet is reached at t = 3 seconds. Completing the square is a powerful method because it provides a clear and direct way to transform the quadratic equation into a form that reveals the vertex, making it easy to identify the maximum or minimum value and the time at which it occurs. This method reinforces algebraic manipulation skills and provides a solid foundation for understanding quadratic functions.

Vertex Formula

The vertex formula is a direct method to find the coordinates of the vertex of a parabola given by the quadratic equation h(t) = at^2 + bt + c. The formula for the t-coordinate (h) of the vertex is h = -b / 2a. Once we find the t-coordinate, we can substitute it back into the original equation to find the height-coordinate (k) of the vertex. For the function h(t) = -16t^2 + 96t + 6, we have a = -16, b = 96, and c = 6. Applying the vertex formula: t = -b / 2a = -96 / (2 * -16) = -96 / -32 = 3. So, the time at which the object reaches its maximum height is 3 seconds. To find the maximum height, substitute t = 3 back into the original equation: h(3) = -16(3)^2 + 96(3) + 6 = -16(9) + 288 + 6 = -144 + 288 + 6 = 150. This confirms that the maximum height is 150 feet, reached at t = 3 seconds. The vertex formula is a straightforward and efficient method for finding the vertex of a parabola. It requires minimal algebraic manipulation and provides a quick solution for the time at maximum height. This method is particularly useful when the primary goal is to find the vertex without needing to rewrite the entire equation in vertex form. Understanding and applying the vertex formula enhances problem-solving skills and provides a valuable tool for analyzing quadratic functions.

Calculus Approach

Calculus provides a powerful approach to finding the maximum or minimum of a function by using derivatives. The derivative of a function gives us the rate of change of the function, and at a maximum or minimum point, the rate of change is zero. For the function h(t) = -16t^2 + 96t + 6, we first find the derivative with respect to t: h'(t) = d/dt (-16t^2 + 96t + 6) = -32t + 96. To find the critical points, we set the derivative equal to zero and solve for t: -32t + 96 = 0. Add 32t to both sides: 96 = 32t. Divide by 32: t = 96 / 32 = 3. So, the critical point occurs at t = 3 seconds. To confirm that this critical point is a maximum, we can use the second derivative test. Find the second derivative: h''(t) = d/dt (-32t + 96) = -32. Since the second derivative is negative (-32), the function has a maximum at t = 3. Substitute t = 3 back into the original equation to find the maximum height: h(3) = -16(3)^2 + 96(3) + 6 = -16(9) + 288 + 6 = -144 + 288 + 6 = 150. This confirms that the maximum height is 150 feet, reached at t = 3 seconds. The calculus approach provides a rigorous method for finding maxima and minima of functions. By using derivatives, we can identify critical points and use the second derivative test to confirm whether they are maxima or minima. This method is widely applicable in various fields, including physics, engineering, and economics, where optimization problems are common. Understanding and applying calculus techniques enhances analytical skills and provides a deeper understanding of function behavior.

By applying the methods discussed—completing the square, using the vertex formula, and employing calculus—we consistently arrive at the same conclusion: the object reaches its maximum height at t = 3 seconds. This result confirms the accuracy and reliability of our methods and provides a clear answer to the problem.

Detailed Explanation

Completing the Square Method:

As demonstrated earlier, completing the square transforms the function into vertex form: h(t) = -16(t - 3)^2 + 150. The vertex of the parabola is (3, 150), indicating that the maximum height is reached at t = 3 seconds.

Vertex Formula Method:

Using the vertex formula t = -b / 2a, we found t = -96 / (2 * -16) = 3 seconds. This direct approach confirms the time at which the maximum height is reached.

Calculus Method:

By finding the first derivative h'(t) = -32t + 96 and setting it to zero, we obtained t = 3 seconds. The negative second derivative h''(t) = -32 confirmed that this point is a maximum.

The consistency across these methods reinforces our understanding of the problem and the solution. Whether using algebraic manipulation or calculus techniques, the result remains the same: the object reaches its maximum height after 3 seconds. This comprehensive approach not only solves the problem but also provides a deeper insight into the underlying mathematical principles.

In summary, we have analyzed the function h(t) = -16t^2 + 96t + 6, which represents the height of an object projected into the air. By using multiple methods—completing the square, applying the vertex formula, and employing calculus—we determined that the object reaches its maximum height of 150 feet after 3 seconds. This problem illustrates the power of mathematical tools in solving real-world scenarios, particularly in physics and engineering.

The ability to analyze projectile motion is crucial in various applications, from sports to military science. Understanding quadratic functions and their properties allows us to predict and optimize the trajectories of objects in flight. The methods we have explored not only provide solutions but also enhance our problem-solving skills and mathematical intuition. By mastering these techniques, we can tackle more complex problems and gain a deeper appreciation for the elegance and utility of mathematics. The consistency of the results obtained through different methods underscores the robustness of mathematical principles. Whether using algebraic techniques or calculus, the underlying concepts provide a solid foundation for understanding and solving problems related to projectile motion and other quadratic phenomena.

The final answer is B. 3 seconds.