/*! 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 19 Exercises \(18-20\) deal with th... [FREE SOLUTION] | 91Ó°ÊÓ

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Exercises \(18-20\) deal with the path of a projectile (such as a baseball, a rocket, or an arrow. If the projectile is fired with an initial velocity of \(v\) feet per second at angle of \(t\) radians and its initial height is k feet, then the path of the projectile is given by $$ y=\left(\frac{-16}{v^{2}} \sec ^{2} t\right) x^{2}+(\tan t) x+k $$ You can think of the projectile as being fired in the direction of the \(x\) -axis from the point ( \(0, k\) ) on the \(y\) -axis. Do Exercise 18 for a projectile that is fired from ground level at an initial velocity of 80 feet per second at an angle of .4 radians.

Short Answer

Expert verified
Answer: The equation for the path of the projectile is approximately \(y = -0.00333 x^{2} + 0.4228x\).

Step by step solution

01

Identify the given values in the problem

We are given the values: Initial Velocity (\(v\)) = 80 feet per second Angle (\(t\)) = 0.4 radians Initial Height (\(k\)) = 0 feet
02

Plug the given values into the formula

Now we need to substitute the given values into the formula: $$ y=\left(\frac{-16}{v^{2}} \sec ^{2} t\right) x^{2}+(\tan t) x+k $$
03

Find \(\sec^2(t)\) and \(\tan(t)\)

Calculate \(\sec^2(0.4)\) and \(\tan(0.4)\) using a calculator or trigonometric table: \(\sec^2(0.4) = \frac{1}{\cos^2(0.4)} \approx 1.3303\) \(\tan(0.4) \approx 0.4228\)
04

Substitute the trigonometric values into the formula

Now, substitute the calculated values of \(\sec^2(0.4)\) and \(\tan(0.4)\) into the formula: $$ y=\left(\frac{-16}{80^{2}} \cdot 1.3303 \right) x^{2}+(0.4228) x+0 $$
05

Simplify the equation

After substitution, simplify the formula to find the path of the projectile: $$ y= \left(\frac{-16}{6400} \cdot 1.3303 \right) x^{2} + 0.4228x $$ $$ y \approx -0.00333 x^{2} + 0.4228x $$ The equation for the path of the projectile is: $$ y \approx -0.00333 x^{2} + 0.4228x $$

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Key Concepts

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

Trigonometry in Projectile Motion
In the study of projectile motion in precalculus, trigonometry plays a crucial role as it helps describe the relationship between angles and distances. When analyzing the path of a projectile, such as a baseball or a rocket, trigonometric functions like sine, cosine, and tangent are employed to decipher the motion's components.

For instance, the tangent of the angle of projection, represented by \(\tan(t)\), allows us to determine the linear relationship between the projectile's height (\(y\)) and horizontal distance (\(x\)) at any point during its flight. On the other hand, the secant squared function, denoted as \(\sec^2(t)\), which is the reciprocal of the cosine squared (\(\cos^2(t)\)), becomes pivotal when calculating the effects of gravity on the projectile's trajectory.

These trigonometric functions, when inserted into the equation of a projectile's path, make it possible to predict the height at any given horizontal distance. This application is not only vital in academics but is also practical in fields like engineering, aeronautics, and even sports.
Quadratic Functions and Projectile Paths
Quadratic functions often represent the trajectory of objects thrown through the air, known as projectile motion. The path of a projectile, when not accounting for air resistance, follows a parabola, which is the graph of a quadratic function. This is demonstrated by the \( x^2 \) term in the projectile's equation.

Given the standard form \(y = ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants, the quadratic coefficient \(a\) indicates the curvature of the parabola. In the context of projectile motion, this coefficient is influenced by the force of gravity and the projectile's initial velocity and angle of projection. A negative value of \(a\) implies that the parabola opens downward, which is expected as the projectile eventually returns to the ground due to gravity.

By analyzing the quadratic function, we can also identify key features of the motion, such as its maximum height and the distance it travels before landing, by finding the vertex and zeros of the parabola, respectively.
Understanding Initial Velocity in Projectile Motion
In physics and precalculus, when we discuss projectile motion, the initial velocity, denoted \(v\), is the speed at which the projectile is launched. It is a vector quantity, which means it possesses both magnitude and direction. The magnitude is the speed component, and the direction is generally given by the angle of projection.

Initial velocity plays a central role in determining the range, height, and overall trajectory of the projectile. In our example, the initial velocity is 80 feet per second. This value directly influences the impact of gravity represented in the quadratic function's coefficient. The greater the initial velocity, the further and higher the projectile can go before gravity pulls it back down. Conversely, a lower initial velocity limits the projectile's range and peak height.

Students often misunderstand that initial velocity is not just the speed at the start but is critical in defining the entire course of the projectile's flight. Without considering the initial velocity, predictions about the projectile's path are baseless.
Angle of Projection's Role in Projectile Motion
The angle of projection, typically denoted by \(t\) and given in radians or degrees, is the angle at which a projectile is launched with respect to the horizontal axis. This angle is a key factor that works in tandem with the initial velocity to shape the projectile's trajectory.

The ideal angle for achieving the maximum horizontal distance—also known as the range—in perfect conditions devoid of air resistance, is 45 degrees or \(\frac{\pi}{4}\) radians. Launch angles greater or less than this angle result in a shorter range. Moreover, the launch angle also affects the height or the maximum vertical distance the projectile can reach. A higher angle increases the vertical component of the initial velocity, leading to a higher but shorter path, while a lower angle favors distance over height.

In our exercise, the angle of projection is 0.4 radians. This angle is smaller than 45 degrees, suggesting that the emphasis is more on the range rather than the peak height of the projectile. Understanding the relationship between the angle, the initial velocity, and the effects of gravity is essential for accurately predicting the path of a projectile.

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Most popular questions from this chapter

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The volume \(V(t)\) of air (in cubic inches) in an adult's lungs \(t\) seconds after exhaling is approximately $$V(t)=55+24.5 \sin \left(\frac{\pi x}{2}-\frac{\pi}{2}\right)$$ (a) Find the maximum and minimum amount of air in the lungs. (b) How often does the person exhale? (c) How many breaths per minute does the person take?

Explore various ways in which a calculator can produce inaccurate graphs of trigonometric functions. These exercises also provide examples of two functions, with different graphs, whose graphs appear identical in certain viewing windows. Choose a viewing window with \(-3 \leq y \leq 3\) and \(0 \leq x \leq k\) where \(k\) is chosen as follows. $$\begin{array}{|l|c|} \hline \text { Width of Screen } & k \\ \hline \begin{array}{l} \text { 95 pixels } \\ (\mathrm{TI}-83 / 84+) \end{array} & 188 \pi \\ \hline \begin{array}{l} \text { 127 pixels } \\ \text { (TI-86, Casio) } \end{array} & 252 \pi \\ \hline \begin{array}{l} \text { 131 pixels } \\ \text { (HP-39gs) } \end{array} & 260 \pi \\ \hline \begin{array}{l} \text { 159 pixels } \\ \text { (TI-89) } \end{array} & 316 \pi \\ \hline \end{array}$$ (a) Graph \(y=\cos x\) and the constant function \(y=1\) on the same screen. Do the graphs look identical? Are the functions the same? (b) Use the trace feature to move the cursor along the graph of \(y=\cos x,\) starting at \(x=0 .\) For what values of \(x\) did the calculator plot points? [Hint: \(2 \pi \approx 6.28 .]\) Use this information to explain why the two graphs look identical.

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