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Chapter 2: Problem 35

A stone is thrown straight up from the roof of an 80 -ft building. The distance of the stone from the ground at any time \(t\) (in seconds) is given by $$ h(t)=-16 t^{2}+64 t+80 $$ a. Sketch the graph of \(h\). b. At what time does the stone reach the highest point? What is the stone's maximum height from the ground?

Short Answer

Expert verified
The stone reaches the highest point at \(t=2\) seconds with a maximum height of 144 ft from the ground. The graph of \(h(t)\) is a downward-opening parabola with vertex at (2,144).

Step by step solution

01

Determine the time at which the stone reaches the maximum height

To find the time when the stone reaches the maximum height, we need to calculate the vertex of the parabola: \((t_v, h(t_v))\), where \(t_v\) is the x-coordinate of the vertex. Recall that for a quadratic function in the form \(f(x) = ax^2 + bx + c\), the x-coordinate of the vertex (in our case, \(t_v\)) can be found using the formula: \(t_v = -\frac{b}{2a}\) For our function, \(h(t) = -16t^2 + 64t + 80\), we have: - a = -16 - b = 64 - c = 80 Now we can find \(t_v\): \(t_v = -\frac{64}{2(-16)} = -\frac{64}{-32} = 2\) Thus, the stone reaches the highest point at 2 seconds.
02

Calculate the maximum height from the ground

Now that we know the time at which the stone reaches the highest point, we can find the maximum height by plugging this value into our function h(t): \(h(2) = -16(2)^2 + 64(2) + 80 = -16(4) + 128 + 80 = 144\) The maximum height from the ground is 144 ft.
03

Sketch the graph of h(t)

To sketch the graph, we need the vertex, which we found in steps 1 and 2: (2, 144) - the highest point of the parabola or the maximum height at t = 2 seconds. Also, the parabola is "concave down" since the coefficient of \(t^2\) is negative (-16). We know that the stone starts at an elevation of 80 ft when t = 0: \(h(0) = -16(0)^2 + 64(0) + 80 = 80\) To easily draw the parabola, we can find one more point on our graph: Let's check the height at t = 1 second: \(h(1) = -16(1)^2 + 64(1) + 80 = -16 + 64 + 80 = 128\) We have the following points: - (0, 80): starting point - (1, 128): the height at t = 1 second - (2, 144): the maximum height Now we can sketch the graph of the function \(h(t)\) using these reference points, knowing that the parabola is "concave down": 1. Draw the t-axis (representing the time in seconds) and the h(t)-axis (representing the height from the ground). 2. Mark the points (0, 80), (1, 128), and (2, 144) on the graph. 3. Draw a parabola (concave down) that passes through these points, with the vertex at (2, 144), representing the highest point. To conclude: a) We have sketched the graph of the function h(t). b) The stone reaches the highest point at 2 seconds, with a maximum height of 144 ft from the ground.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Parabola Sketching
When sketching a parabola, it's important to identify certain key features. For a quadratic function like \(h(t) = -16t^2 + 64t + 80\), we start by observing the general shape dictated by the coefficient of \(t^2\). Here, since the coefficient \(-16\) is negative, the parabola opens downwards or is concave down. This indicates that the graph will have a "hill" shape, with a distinct highest point, which is the vertex.

Steps to Sketch a Parabola

  • Identify the axis of symmetry, which runs vertically through the vertex. For our function, \(t = 2\) is this axis, dividing the parabola into two symmetrical halves.
  • Find key points: starting with the vertex (2, 144), and additional points like (0, 80) and (1, 128) as calculated.
  • Draw the t-axis (horizontal axis for time) and the h(t)-axis (vertical axis for height), marking out these points.
  • Using these points, sketch the curve showing a smooth "U" turned upside down.
Visualizing the trajectory of the stone through this graph gives a better understanding of how it ascends and eventually descends after hitting its peak.
Vertex of a Parabola
The vertex is a crucial part of any parabola, especially in problems involving maximum or minimum values, like in projectile motion. It serves as the parabola's turning point. For a quadratic function in the standard form \(ax^2 + bx + c\), the vertex can be calculated using the formula \(t_v = -\frac{b}{2a}\). Here, \(t_v\) gives us the time at which the peak or low point occurs. For the function \(h(t) = -16t^2 + 64t + 80\):
  • a = -16, b = 64, and c = 80.
  • Therefore, \(t_v = -\frac{64}{2(-16)} = 2\).
  • This tells us that the vertex occurs when \(t = 2\).
We also define the vertex's corresponding height, \(h(t_v)\), by substituting \(t_v\) back into the height equation, yielding \(h(2) = 144\) feet. Thus, the vertex here is (2, 144), indicating the maximum height the stone reaches in this model.
Maximum Height Calculation
Calculating the maximum height is an integral part of analyzing projectile motion. It involves determining the vertex of the parabola, as this is the highest point. Using our example, the vertex of the quadratic function \(h(t) = -16t^2 + 64t + 80\), found earlier, is (2, 144).

How to Calculate the Maximum Height

  • First, determine the time when this maximum occurs using the vertex formula \(t_v = -\frac{b}{2a}\).
  • Plug \(t_v = 2\) into the original equation to get the height: \(h(2) = -16(2)^2 + 64(2) + 80 = 144\).
  • This process yields the maximum height of 144 feet.
This not only gives the absolute highest point reached during the stone's trajectory but also aids in understanding the motion path. By analyzing this, one can further extend the application to other real-world problems, ensuring a better grasp of quadratic motion.

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