Rocket Maximum Height Calculation A Physics Problem

Hey guys! Let's dive into an interesting physics problem involving a rocket launch. We're going to figure out the maximum height a rocket reaches when launched straight up into the air. We've got the initial conditions and the equation that describes the rocket's height over time, so let's get started!

In this article, we'll break down how to calculate the maximum height of a rocket given its initial velocity, initial height, and the equation of motion. This involves a bit of algebra and calculus, but don't worry, we'll take it step by step. Understanding these concepts is crucial not only for solving physics problems but also for grasping real-world applications like space travel and engineering. So, let's launch into it and see how high we can go!

The problem we're tackling today involves a rocket launched vertically into the air. The rocket starts with an initial velocity of 10 feet per second (ft/sec) and an initial height of 250 feet. The height of the rocket at any given time t is described by the equation:

h(t) = -16t^2 + 10t + 250

Where:

• h(t) is the height of the rocket at time t (in feet).
• t is the time elapsed since launch (in seconds).

The goal is to determine the maximum height the rocket reaches. This is a classic problem in physics that combines concepts from kinematics and calculus. By understanding how to solve this, we can apply these principles to other similar problems involving projectile motion and optimization.

The key question we're trying to answer is: What is the maximum height the rocket achieves during its flight? To find this, we'll need to use the given equation and some mathematical techniques to identify the highest point in the rocket's trajectory.

The equation h(t) = -16t^2 + 10t + 250 is a quadratic equation, which represents a parabola when graphed. In this context, the parabola describes the path of the rocket as it goes up and comes back down due to gravity. Let's break down each term in the equation:

• -16t²: This term represents the effect of gravity on the rocket. The negative sign indicates that gravity is pulling the rocket downwards. The coefficient -16 is related to the acceleration due to gravity (approximately -32 ft/sec²), with -16 being half of this value, as it appears in the height equation.
• 10t: This term represents the initial upward velocity of the rocket. The coefficient 10 is the initial velocity (10 ft/sec), and multiplying it by time t gives the distance the rocket would travel upwards if there were no gravity.
• 250: This constant term represents the initial height of the rocket. It's the height from which the rocket was launched at time t = 0.

The parabolic shape of the rocket's trajectory means it will reach a maximum height at its vertex. The vertex is the highest point on the parabola, and it's the point where the rocket momentarily stops moving upwards before it starts to descend. Finding the vertex of this parabola will give us both the time at which the rocket reaches its maximum height and the maximum height itself.

One way to find the maximum height is by completing the square. This method transforms the quadratic equation into vertex form, which directly reveals the coordinates of the vertex (and hence the maximum height). Let's walk through the steps:

1. Rewrite the equation: Start with the given equation:

   h(t) = -16t^2 + 10t + 250

2. Factor out -16 from the first two terms: This isolates the t² and t terms:

   h(t) = -16(t^2 - (10/16)t) + 250

3. Simplify the fraction: Reduce 10/16 to 5/8:

   h(t) = -16(t^2 - (5/8)t) + 250

4. Complete the square: To complete the square, we need to add and subtract (b/2)² inside the parenthesis, where b is the coefficient of the t term (-5/8 in this case). So, (b/2)² = (-5/16)² = 25/256:

   h(t) = -16(t^2 - (5/8)t + 25/256 - 25/256) + 250

5. Rewrite as a perfect square: The terms inside the parenthesis now form a perfect square:

   h(t) = -16((t - 5/16)^2 - 25/256) + 250

6. Distribute -16 and simplify: Multiply -16 through the parenthesis and simplify:

   h(t) = -16(t - 5/16)^2 + 16*(25/256) + 250

   h(t) = -16(t - 5/16)^2 + 25/16 + 250

7. Combine constants: Add the constants together:

   h(t) = -16(t - 5/16)^2 + (25 + 4000) / 16

   h(t) = -16(t - 5/16)^2 + 4025/16

Now the equation is in vertex form: h(t) = a(t - h)^2 + k, where (h, k) is the vertex of the parabola. In this case, the vertex is (5/16, 4025/16). Therefore, the maximum height is 4025/16 feet.

Another method to find the maximum height involves using calculus. The maximum height occurs at the vertex of the parabola, which corresponds to the point where the derivative of the height function is zero. Here’s how we can use calculus to solve this:

1. Find the derivative of h(t): The derivative, h'(t), gives the instantaneous rate of change of the rocket's height with respect to time (i.e., its vertical velocity). Differentiate the original equation:

   h(t) = -16t^2 + 10t + 250

   h'(t) = -32t + 10

2. Set the derivative to zero and solve for t: The maximum height occurs when the rocket's vertical velocity is zero. This is because the rocket momentarily stops moving upwards before it starts falling back down. Set h'(t) to zero and solve for t:

   -32t + 10 = 0

   32t = 10

   t = 10/32

   t = 5/16

   This value of t is the time at which the rocket reaches its maximum height.

3. Substitute t back into h(t): Plug the value of t = 5/16 back into the original height equation to find the maximum height:

   h(5/16) = -16(5/16)^2 + 10(5/16) + 250

   h(5/16) = -16(25/256) + 50/16 + 250

   h(5/16) = -25/16 + 50/16 + 250

   h(5/16) = 25/16 + 250

   h(5/16) = (25 + 4000) / 16

   h(5/16) = 4025/16

So, the maximum height the rocket reaches is 4025/16 feet, which matches the result we obtained using the completing the square method.

A more direct way to find the vertex of a parabola given by the quadratic equation h(t) = at² + bt + c is to use the vertex formula. The t-coordinate of the vertex (which represents the time at which the maximum height is reached) is given by:

t_vertex = -b / (2a)

And the maximum height (the h-coordinate of the vertex) can be found by substituting this t_vertex back into the original equation.

Let's apply this to our equation:

h(t) = -16t^2 + 10t + 250

Here, a = -16, b = 10, and c = 250.

1. Find the time at which the vertex occurs: Use the vertex formula to find t_vertex:

   t_vertex = -b / (2a) = -10 / (2 * -16) = -10 / -32 = 5/16

2. Substitute t_vertex into h(t): Plug t_vertex = 5/16 back into the original equation to find the maximum height:

   h(5/16) = -16(5/16)^2 + 10(5/16) + 250

   h(5/16) = -16(25/256) + 50/16 + 250

   h(5/16) = -25/16 + 50/16 + 250

   h(5/16) = 25/16 + 250

   h(5/16) = (25 + 4000) / 16

   h(5/16) = 4025/16

Again, we find that the maximum height is 4025/16 feet. This method provides a quick and efficient way to find the maximum or minimum value of any quadratic function.

Alright guys, we've tackled this rocket launch problem using three different methods: completing the square, calculus, and the vertex formula. Each method led us to the same answer, which reinforces the correctness of our solution. So, the final answer is:

The rocket's maximum height is 4025/16 feet, which is equal to 251.5625 feet. Remember, we didn't round the answer, as requested in the problem statement.

We've successfully calculated the maximum height of a rocket launched into the air using a quadratic equation. By understanding the equation and applying techniques like completing the square, calculus, and the vertex formula, we can solve various optimization problems. I hope you found this exploration insightful and helpful! Keep exploring the exciting world of physics, and who knows, maybe you'll be designing rockets one day!