/*! 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 8 A baseball is hit 2 feet above h... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A baseball is hit 2 feet above home plate, and the position of the ball \(t\) seconds later is \(\mathbf{r}(t)=\left\langle 40 t,-16 t^{2}+31 t+2\right\rangle\) ft. Find each of the following values. a. The time of flight of the baseball b. The range of the baseball

Short Answer

Expert verified
Question: Determine the time of flight and the range of a baseball that follows the position vector \(\mathbf{r}(t)=\left\langle 40 t,-16 t^{2}+31 t+2\right\rangle\). Answer: To find the time of flight and range, first identify the position vector components: \(\mathbf{r_x}(t) = 40 t\) \(\mathbf{r_y}(t) = -16 t^{2} + 31 t + 2\) Next, find the time of flight by solving for when \(\mathbf{r_y}(t) = 0\): \(t_{flight} = \frac{-31 + \sqrt{31^2 + 128}}{-32}\) Finally, calculate the range of the baseball by plugging the time of flight into the horizontal component of the position vector: \(\mathbf{r_x}(t_{flight}) = 40 \times t_{flight}\)

Step by step solution

01

Identify the position vector components

The position vector of the baseball is given by \(\mathbf{r}(t)=\left\langle 40 t,-16 t^{2}+31 t+2\right\rangle\). This means that the horizontal component \(\mathbf{r_x}(t)\) and the vertical component \(\mathbf{r_y}(t)\) are given by: \(\mathbf{r_x}(t) = 40 t\) \(\mathbf{r_y}(t) = -16 t^{2} + 31 t + 2\)
02

Find the time of flight

To find the time of flight, we need to find the time at which the vertical position of the ball reaches the ground again; in other words, when \(\mathbf{r_y}(t) = 0\). Solve the equation \(-16 t^{2} + 31 t+ 2 = 0\) for \(t\). This is a quadratic equation of the form \(ax^2 + bx + c = 0\), with \(a = -16\), \(b = 31\), and \(c = 2\). We can use the quadratic formula to find its solutions: \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) Using the values of \(a\), \(b\), and \(c\), we get: \(t = \frac{-31 \pm \sqrt{31^2 - 4(-16)(2)}}{-32}\) Solving this equation, we get two possible values for \(t\): one positive and one negative. Because time cannot be negative, we can ignore the negative solution and take the positive solution as the time of flight.
03

Calculate the range of the baseball

Now that we have found the time of flight, we can find the range of the baseball by determining the horizontal distance when the ball hits the ground. To do this, plug the time of flight found in step 2 into the horizontal component of the position vector: \(\mathbf{r_x}(t_{flight}) = 40 \times t_{flight}\) Calculate the value and we will get the range of the baseball.

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 91Ó°ÊÓ!

Key Concepts

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

Position Vector Components
Understanding the position vector components is crucial when analyzing the motion of a projectile, like a baseball. In the given exercise, the position of the baseball as a function of time is represented by a vector \( \mathbf{r}(t) \). The two parts of this vector, \( \mathbf{r_x}(t) \) and \( \mathbf{r_y}(t) \), represent the horizontal and vertical positions of the baseball, respectively.

The horizontal component \( \mathbf{r_x}(t) = 40t \) tells us that for every second that passes, the baseball travels 40 feet horizontally, which shows a constant speed. On the other hand, the vertical component \( \mathbf{r_y}(t) = -16t^2 + 31t + 2 \) is more complex because it includes a quadratic term, indicative of the acceleration due to gravity, as well as a linear term and a constant. The constant represents the initial height from which the baseball is hit. By analyzing these components, we learn not only where the baseball is at any given moment but also information about its speed and acceleration in both directions.
Quadratic Equations
When dealing with the vertical motion of a projectile, we often encounter a quadratic equation. This type of equation can be written in the standard form \( ax^2 + bx + c = 0 \) and is solved using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a \) represents the coefficient of \( t^2 \) (reflecting acceleration), \( b \) is the coefficient of \( t \) (initial velocity), and \( c \) is the constant term (initial height).

In our baseball scenario, we used this formula to find the time at which the baseball returns to the height of 2 feet—the point at which it was hit. The positive solution to this quadratic equation gave us the time of flight of the baseball. It is essential for students to understand the discriminant part \( \sqrt{b^2 - 4ac} \) since it shows the nature of the roots of the equation - in the case of projectile motion, there's typically one positive root that is physically meaningful.
Horizontal and Vertical Motion
Breaking down the motion of a projectile into horizontal and vertical components simplifies its analysis. For the horizontal motion, we assume a constant velocity because, in the absence of air resistance, no horizontal forces are acting on the baseball. Thus, the horizontal displacement can be found by multiplying the velocity with the time of flight.

Contrastingly, the vertical motion is influenced by gravity and is therefore represented by a quadratic equation, as seen in the vertical component of the position vector. The interplay between the initial vertical velocity and the acceleration due to gravity determines the shape of the projectile's path in the vertical plane. When the vertical displacement becomes zero, we can find the time at which the baseball hits the ground, which we then use to calculate the horizontal range.
Kinematics
The study of kinematics is concerned with how objects move—speed, velocity, and acceleration—without considering the forces causing the motion. It is an essential concept for understanding projectile motion, such as our baseball example. Kinematic equations allow us to relate the various parameters such as initial velocity, final velocity, acceleration, time, and displacement.

In solving the baseball problem, we strictly adhere to kinematic principles. The position vector \( \mathbf{r}(t) \) uses kinematic equations to express the baseball's location at any time \( t \). The horizontal motion, showing constant speed, and the vertical motion, affected by gravity, are both parts of the complete kinematic description of the baseball's flight. Familiarity with kinematic concepts is instrumental in accurately predicting the behavior of moving objects and is fundamental for students tackling physics problems involving motion.

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

Relationship between \(\mathbf{T}, \mathbf{N},\) and a Show that if an object accelerates in the sense that \(\frac{d^{2} s}{d t^{2}}>0\) and \(\kappa \neq 0,\) then the acceleration vector lies between \(\mathbf{T}\) and \(\mathbf{N}\) in the plane of \(\mathbf{T}\) and \(\mathbf{N}\). Show that if an object decelerates in the sense that \(\frac{d^{2} s}{d t^{2}}<0,\) then the acceleration vector lies in the plane of \(\mathbf{T}\) and \(\mathbf{N},\) but not between \(\mathbf{T}\) and \(\mathbf{N} .\)

Solving equations of motion Given an acceleration vector, initial velocity \(\left\langle u_{0}, v_{0}, w_{0}\right\rangle,\) and initial position \(\left\langle x_{0}, y_{0}, z_{0}\right\rangle,\) find the velocity and position vectors, for \(t \geq \mathbf{0}\). $$a(t)=\langle 0,0,10\rangle,\left\langle u_{0}, v_{0}, w_{0}\right\rangle=\langle 1,5,0\rangle, \left\langle x_{0}, y_{0}, z_{0}\right\rangle=\langle 0,5,0\rangle$$

Tilted ellipse Consider the curve \(\mathbf{r}(t)=\langle\cos t, \sin t, c \sin t\rangle,\) for \(0 \leq t \leq 2 \pi,\) where \(c\) is a real number. Assuming the curve lies in a plane, prove that the curve is an ellipse in that plane.

Consider a particle that moves in a plane according to the function \(\mathbf{r}(t)=\left\langle\sin t^{2}, \cos t^{2}\right\rangle\) with an initial position (0,1) at \(t=0\) a. Describe the path of the particle, including the time required to return to the initial position. b. What is the length of the path in part (a)? c. Describe how the motion of this particle differs from the motion described by the equations \(x=\sin t\) and \(y=\cos t\) d. Consider the motion described by \(x=\sin t^{n}\) and \(y=\cos t^{n}\) where \(n\) is a positive integer. Describe the path of the particle, including the time required to return to the initial position. e. What is the length of the path in part (d) for any positive integer \(n ?\) f. If you were watching a race on a circular path between two runners, one moving according to \(x=\sin t\) and \(y=\cos t\) and one according to \(x=\sin t^{2}\) and \(y=\cos t^{2},\) who would win and when would one runner pass the other?

Consider the following curves. a. Graph the curve. b. Compute the curvature. c. Graph the curvature as a ficnction of the parameter. d. Identify the points (if any) at which the curve has a maximum or minimum curvature. e. Verify that the graph of the curvature is consistent with the graph of the curve. $$\mathbf{r}(t)=\left\langle t, t^{2}\right\rangle, \text { for }-2 \leq t \leq 2 \quad \text { (parabola) }$$

See all solutions

Recommended explanations on Math Textbooks

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.