/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 39 A shot-put athlete releases the ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 39

A shot-put athlete releases the shot at a speed of 14 meters per second, at an angle of 45 degrees to the horizontal (ground level). The height \(y\) (in meters above the ground) of the shot is given by the function $$ y=2+x-\frac{1}{20} x^{2} $$ where \(x\) is the horizontal distance the shot has traveled (in meters). a. What was the height of the shot at the moment of release? b. How high is the shot after it has traveled 4 meters horizontally from the release point? 16 meters? c. Find the highest point reached by the shot in its flight. d. Draw a sketch of the height of the shot and indicate how far the shot is from the athlete when it lands.

Short Answer

Expert verified
a. 2 meters b. 5.2 meters c. 7 meters d. 21.8 meters

Step by step solution

01

Understand the given function

The height of the shot-put, in meters above the ground, is given by the function: \[ y = 2 + x - \frac{1}{20} x^{2} \] In this function, \(x\) represents the horizontal distance in meters.
02

Find the height at the moment of release

At the moment of release, the horizontal distance \(x\) is 0 meters. Substitute \(x = 0\) into the function: \[ y = 2 + 0 - \frac{1}{20}(0)^{2} = 2 \] So, the height of the shot at the moment of release is 2 meters.
03

Calculate the height at \(x = 4\) meters

Substitute \(x = 4\) into the function: \[ y = 2 + 4 - \frac{1}{20}(4)^{2} = 2 + 4 - \frac{1}{20}(16) = 2 + 4 - \frac{16}{20} = 6 - 0.8 = 5.2 \] Therefore, the height is 5.2 meters after traveling 4 meters horizontally.
04

Calculate the height at \(x = 16\) meters

Substitute \(x = 16\) into the function: \[ y = 2 + 16 - \frac{1}{20}(16)^{2} = 2 + 16 - \frac{1}{20}(256) = 18 - 12.8 = 5.2 \] So, the height is 5.2 meters after traveling 16 meters horizontally.
05

Find the highest point of the shot

To find the highest point, we need to find the vertex of the parabola. The function is in the standard form \(y = ax^2 + bx + c\), where \(a = -\frac{1}{20}, b = 1, c = 2\). The vertex is at \(x = -\frac{b}{2a}\). Substitute the known values: \[ x = -\frac{1}{2(-\frac{1}{20})} = \frac{1}{\frac{1}{10}} = 10 \] Now, substitute \(x = 10\) into the function to find the height: \[ y = 2 + 10 - \frac{1}{20}(10)^{2} = 2 + 10 - \frac{1}{20}(100) = 2 + 10 - 5 = 7 \] The highest point reached by the shot is 7 meters.
06

Find the distance when the shot lands

The shot lands when the height \(y = 0\). Set the function equal to zero and solve for \(x\): \[ 0 = 2 + x - \frac{1}{20} x^{2} \] Rearrange into a quadratic equation: \[ 0 = -\frac{1}{20} x^{2} + x + 2 \] Multiply by 20 to clear the fraction: \[ 0 = -x^{2} + 20x + 40 \] Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), with \(a = -1, b = 20, c = 40\): \[ x = \frac{20 \pm \sqrt{400 + 160}}{-2} = \frac{20 \pm \sqrt{560}}{-2} \] \[ x = \frac{20 \pm 2\sqrt{140}}{-2} = 10 \pm \sqrt{140} \] Only the positive root is meaningful; hence: \[ x = 10 + \sqrt{140} \approx 21.8 \] The shot lands approximately 21.8 meters from the athlete.
07

Draw the sketch

Draw the height function \(y = 2 + x - \frac{1}{20} x^{2}\) on coordinate axes. Mark important points such as the release point \((0, 2)\), maximum height \((10, 7)\), points at \(x = 4\) and \(x = 16\), and the landing point \((21.8, 0)\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Quadratic equations
A quadratic equation is a type of polynomial equation of degree 2. It has the general form: \[ ax^2 + bx + c = 0 \]where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\). In our shot-put problem, the height function is a quadratic equation where the horizontal distance forms the quadratic term. Quadratic equations can describe the path of projectiles, like our shot-put, showing a parabolic trajectory.The solutions of quadratic equations can be found using various methods, such as factoring, completing the square, or the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]This formula is especially useful for finding the points where a quadratic function crosses the x-axis, which in the context of the shot-put problem, helps determine the distance at which the shot lands. Understanding how to solve these equations is crucial for finding key points in projectile motion.
Vertex of a parabola
The vertex of a parabola is a significant feature; it represents the highest or lowest point on the graph. For a parabolic trajectory of a shot-put, the vertex gives us the maximum height reached by the shot. The general form of a quadratic function \(ax^2 + bx + c\) simplifies finding the vertex. The x-coordinate of the vertex is given by:\[ x = -\frac{b}{2a} \]Substituting this value back into the quadratic function determines the y-coordinate, which is the maximum height in our problem. In our example, the highest point of the shot-put occurs when \(x = 10\), leading us to a maximum height of 7 meters. Knowing the vertex helps in graphing and understanding the dynamics of projectile motion.
Height function
A height function describes how the height of a projectile changes with horizontal distance. In the shot-put problem, the height function is:\[ y = 2 + x - \frac{1}{20} x^2 \]Here, \( y \) represents the height in meters above the ground, and \( x \) is the horizontal distance in meters. The function integrates initial conditions and physics of motion to give a parabolic shape. By substituting different values of \( x \) into the function, we can find the height at various points of the shot-put's flight.For example, at the release point (\( x = 0 \)), the height is 2 meters. When the shot-put has traveled 4 meters horizontally, substituting \( x = 4 \) gives a height of 5.2 meters. The function provides a comprehensive way to track the vertical position throughout the projectile's travel.
Graphing functions
Graphing functions helps us visualize the behavior of a projectile. To graph the height function of our shot-put, plot points for different values of \( x \). Key points include where the shot-put is released, reaches its maximum height, and lands. Here's the height function again:\[ y = 2 + x - \frac{1}{20} x^2 \]To graph this:
  • Start by plotting the release point: \( (0, 2) \).
  • Mark the maximum height (vertex): \( (10, 7) \).
  • Include other points, such as \( (4, 5.2) \) and \( (16, 5.2) \).
  • Find where the parabola intersects the x-axis (when \( y = 0 \)). This gives the landing point, approximately \( (21.8, 0) \).
Connecting these points will show the parabolic path. This graph helps to predict and understand projectile motion, ensuring that any calculations or real-world applications are accurately reflected in a visual format.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Polynomial expressions of the form \(a n^{3}+b n^{2}+c n+d\) can be used to express positive integers, such as \(4573,\) using different powers of 10: $$ 4573=4 \cdot 1000+5 \cdot 100+7 \cdot 10+3 $$ or $$ 4573=4 \cdot 10^{3}+5 \cdot 10^{2}+7 \cdot 10^{1}+3 \cdot 10^{0} $$ or if \(n=10\) $$ \begin{aligned} 4573 &=4 \cdot n^{3}+5 \cdot n^{2}+7 \cdot n^{1}+3 \cdot n^{0} \\ &=4 n^{3}+5 n^{2}+7 n+3 \end{aligned} $$ Notice that in order to represent any positive number, the coefficient multiplying each power of 10 must be an integer between 0 and 9 . a. Express 8701 as a polynomial in \(n\) assuming \(n=10\). b. Express 239 as a polynomial in \(n\) assuming \(n=10\). Computers use a similar polynomial system, called binary numbers, to represent numbers as sums of powers of \(2 .\) The number 2 is used because each minuscule switch in a computer can have one of two states, on or off; the symbol 0 signifies off, and the symbol 1 signifies on. Each binary number is built up from a row of switch positions each set at 0 or 1 as multipliers for different powers of 2 . For instance, in the binary number system 13 is represented as 1101 , which stands for $$ \begin{array}{l} 1 \cdot 2^{3}+1 \cdot 2^{2}+0 \cdot 2^{1}+1 \cdot 2^{0}= \\ 1 \cdot 8+1 \cdot 4+0 \cdot 2+1 \cdot 1= \\ 13 \end{array} $$ c. What number does the binary notation 11001 represent? d. Find a way to write 35 as the sum of powers of 2 ; then give the binary notation.

(Graphing program required.) At low speeds an automobile engine is not at its peak efficiency; efficiency initially rises with speed and then declines at higher speeds. When efficiency is at its maximum, the consumption rate of gas (measured in gallons per hour) is at a minimum. The gas consumption rate of a particular car can be modeled by the following equation, where \(G\) is the gas consumption rate in gallons per hour and \(M\) is speed in miles per hour: \(G=0.0002 M^{2}-0.013 M+1.07\) a. Construct a graph of gas consumption rate versus speed. Estimate the minimum gas consumption rate from your graph and the speed at which it occurs. b. Using the equation for \(G,\) calculate the speed at which the gas consumption rate is at its minimum. What is the minimum gas consumption rate? c. If you travel for 2 hours at peak efficiency, how much gas will you use and how far will you go? d. If you travel at \(60 \mathrm{mph}\), what is your gas consumption rate? How long does it take to go the same distance that you calculated in part (c)? (Recall that travel distance = speed \(\times\) time traveled.) How much gas is required for the trip? e. Compare the answers for parts (c) and (d), which tell you how much gas is used for the same-length trip at two different speeds. Is gas actually saved for the trip by traveling at the speed that gives the minimum gas consumption rate? f. Using the function \(G,\) generate data for gas consumption rate measured in gallons per mile by completing the following table. Plot gallons per mile (on the vertical axis) vs. miles per hour (on the horizontal axis). At what speed is gallons per mile at a minimum? g. Add a fourth column to the data table. This time compute miles/gal \(=\mathrm{mph} /(\mathrm{gal} / \mathrm{hr}) .\) Plot miles per gallon vs. miles per hour. At what speed is miles per gallon at a maximum? This is the inverse of the preceding question; we are normally used to maximizing miles per gallon instead of minimizing gallons per mile. Does your answer make sense in terms of what you found for parts (b) and (f)?

a. Construct a quadratic function \(P(s)\) that goes through the point (5,-22) and has two real zeros, one at \(s=-6\) and the other at \(s=4\). b. What is the axis of symmetry? c. What are the coordinates of the vertex? d. What is the vertical intercept?

The exchange rate a bank gave for Canadian dollars on March 2, 2007, was 1.18 Canadian dollars for 1 U.S. dollar. The bank also charges a constant fee of 3 U.S. dollars per transaction. a. Construct a function \(F\) that converts U.S. dollars, \(d,\) to Canadian dollars. b. Construct a function \(G\) that converts Canadian dollars, \(c\), to U.S. dollars. c. What would the function \(F \circ G\) do? Would its input be U.S. or Canadian dollars (i.e., \(d\) or \(c\) )? Construct a formula for \(F \circ G\) d. What would the function \(G \circ F\) do? Would its input be U.S. or Canadian dollars (i.e., \(d\) or \(c\) )? Construct a formula for \(G \circ F\)

Let \((h, k)\) be the coordinates of the vertex of a parabola. Then \(h\) equals the average of the two real zeros of the function (if they exist). For each of the following use this to find \(h,\) and then put the equations into the vertex form, \(y=a(x-h)^{2}+k\) a. A parabola with equation \(y=x^{2}+2 x-8\) b. A parabola with equation \(y=-x^{2}-3 x+4\)
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Applied Mathematics

Read Explanation

Logic and Functions

Read Explanation

Mechanics Maths

Read Explanation

Geometry

Read Explanation

Pure Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.