/*! 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 44 A baseball leaves a pitcher's ha... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A baseball leaves a pitcher's hand horizontally at a speed of \(161 \mathrm{~km} / \mathrm{h} .\) The distance to the batter is \(18.3 \mathrm{~m} .\) (a) How long does the ball take to travel the first half of that distance? (b) The second half? (c) How far does the ball fall freely during the first half? (d) During the second half? (e) Why aren't the quantities in (c) and (d) equal?

Short Answer

Expert verified
(a) 0.2047 s, (b) 0.2047 s, (c) 0.205 m, (d) 0.619 m; (e) falling distance increases with time.

Step by step solution

01

Convert Speed to Meters per Second

First, convert the speed from kilometers per hour to meters per second using the conversion factor: \[1 \text{ km/h} = \frac{1000}{3600} \text{ m/s} = \frac{1}{3.6} \text{ m/s}.\]Hence, \(161 \text{ km/h} = \frac{161}{3.6} \approx 44.72 \text{ m/s}\).
02

Calculate Time for First Half of Distance

The first half of the distance is \(\frac{18.3}{2} \text{ m} = 9.15 \text{ m}\). To find the time it takes to cover this distance, use the formula \(t = \frac{d}{v}\), where \(d\) is distance and \(v\) is velocity:\[t_1 = \frac{9.15}{44.72} \approx 0.2047 \text{ s}.\]
03

Calculate Time for Second Half of Distance

The second half of the distance is also \(9.15 \text{ m}\). Since the speed is constant, the time to travel the second half is the same as the first half: \[t_2 = 0.2047 \text{ s}.\]
04

Calculate Vertical Displacement in First Half

The ball falls freely under gravity during its horizontal motion. Using the formula for the distance fallen under constant acceleration: \[d = \frac{1}{2}gt^2,\]where \(g \approx 9.8 \text{ m/s}^2\) is the acceleration due to gravity and \(t_1 = 0.2047 \text{ s}\): \[d_1 = \frac{1}{2} \times 9.8 \times (0.2047)^2 \approx 0.205 \text{ m}.\]
05

Calculate Vertical Displacement in Second Half

For the total time until the ball reaches the batter, the total time is \(t_1 + t_2 = 2 \times 0.2047 = 0.4094 \text{ s}\). The total vertical displacement is given by:\[d = \frac{1}{2} \times 9.8 \times (0.4094)^2 \approx 0.824 \text{ m}.\]The displacement during the second half is then:\[d_2 = 0.824 - 0.205 = 0.619 \text{ m}.\]
06

Explain Unequal Falling Distances

The ball falls a shorter distance in the first half than in the second because the time under gravity is shorter in the first half. The distance an object falls under uniform acceleration is proportional to the square of the time, so as more time elapses, the falling distance increases more rapidly.

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.

Understanding Kinematics in Projectile Motion
Kinematics is the branch of physics that deals with the motion of objects. It describes how objects move through space over time. In the context of projectile motion, kinematics focuses on two main components: horizontal motion and vertical motion. Each component can be analyzed separately to understand the overall trajectory of the projectile.

In projectile motion, the horizontal and vertical motions are independent of one another. This means that changes in velocity in one direction do not affect the velocity in another direction. In this exercise, the baseball moves horizontally with constant velocity since there is no horizontal acceleration, while the vertical motion is affected by gravity.
This horizontal-vertical separation simplifies the study of projectile motion, as you can tackle each direction individually. The key is to keep in mind that calculations for horizontal motion typically involve constant velocity, while calculations for vertical motion account for gravitational acceleration.
Acceleration Due to Gravity in Projectile Motion
Gravity plays a crucial role in projectile motion by influencing the vertical motion of the object. Specifically, gravity is the force that accelerates the object downward at a constant rate. This rate of acceleration is often denoted by the symbol \(g\) and is approximately equal to \(9.8 \, \text{m/s}^2\) on the surface of the Earth.

In this exercise, the baseball falls freely under the influence of gravity while it moves horizontally towards the batter. This means that while the baseball maintains a constant speed in the horizontal direction, it gains speed in the vertical direction due to gravity.
It's essential to understand that the vertical displacement an object experiences due to gravity increases with time. This is because the longer an object is under the influence of gravity, the faster it moves downward, resulting in a greater distance fallen. This concept explains why the ball falls a longer distance during the second half of its flight compared to the first half, as highlighted in this exercise.
Analyzing Horizontal Motion in Projectile Motion
Horizontal motion in projectile problems like this one involves an object moving with constant velocity, meaning there is no horizontal acceleration. For the baseball, its speed in the horizontal direction remains constant at \(44.72 \, \text{m/s}\), as calculated from the initial speed given in kilometers per hour.

Because there is no horizontal acceleration, we use a straightforward formula to calculate the time the ball takes to cover a certain distance: \(t = \frac{d}{v}\), where \(t\) is time, \(d\) is horizontal distance, and \(v\) is horizontal velocity.
This formula allows us to conclude that the time taken to cover the first half of the distance is exactly the same as for the second half, since the speeds are identical throughout the travel.
  • First half time: \(0.2047\) seconds
  • Second half time: \(0.2047\) seconds
Understanding this aspect of constant horizontal motion helps clarify why the time distributions are equal in both halves, despite differences in how far and fast it falls vertically.

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

Some state trooper departments use aircraft to enforce highway speed limits. Suppose that one of the airplanes has a speed of \(135 \mathrm{mi} / \mathrm{h}\) in still air. It is flying straight north so that it is at all times directly above a north-south highway. A ground observer tells the pilot by radio that a \(70.0 \mathrm{mi} / \mathrm{h}\) wind is blowing but neglects to give the wind direction. The pilot observes that in spite of the wind the plane can travel \(135 \mathrm{mi}\) along the highway in \(1.00 \mathrm{~h}\). In other words, the ground speed is the same as if there were no wind. (a) From what direction is the wind blowing? (b) What is the heading of the plane; that is, in what direction does it point?

A soccer ball is kicked from the ground with an initial speed of \(19.5 \mathrm{~m} / \mathrm{s}\) at an upward angle of \(45^{\circ} .\) A player \(55 \mathrm{~m}\) away in the direction of the kick starts running to meet the ball at that instant. What must be his average speed if he is to meet the ball just before it hits the ground?

A particle \(P\) travels with constant speed on a circle of radius \(r=3.00 \mathrm{~m}\) (Fig. \(4-56\) ) and completes one revolution in \(20.0 \mathrm{~s}\). The particle passes through \(O\) at time \(t=0 .\) State the following vectors in magnitudeangle notation (angle relative to the positive direction of \(x\) ). With respect to \(O,\) find the particle's position vector at the times \(t\) of (a) \(5.00 \mathrm{~s}\), (b) \(7.50 \mathrm{~s},\) and (c) \(10.0 \mathrm{~s}\). (d) For the \(5.00 \mathrm{~s}\) interval from the end of the fifth second to the end of the tenth second, find the particle's displacement. For that interval, find (e) its average velocity and its velocity at the (f) beginning and (g) end. Next, find the acceleration at the (h) beginning and (i) end of that interval.

The velocity \(\vec{v}\) of a particle moving in the \(x y\) plane is given by \(\vec{v}=\left(6.0 t-4.0 t^{2}\right) \hat{\mathrm{i}}+8.0 \hat{\mathrm{j}},\) with \(\vec{v}\) in meters per second and \(t(>0)\) in seconds. (a) What is the acceleration when \(t=3.0 \mathrm{~s} ?\) (b) When (if ever) is the acceleration zero? (c) When (if ever) is the velocity zero? (d) When (if ever) does the speed equal \(10 \mathrm{~m} / \mathrm{s} ?\)

When a large star becomes a supernova, its core may be compressed so tightly that it becomes a neutron star, with a radius of about \(20 \mathrm{~km}\) (about the size of the San Francisco area). If a neutron star rotates once every second, (a) what is the speed of a particle on the star's equator and (b) what is the magnitude of the particle's centripetal acceleration? (c) If the neutron star rotates faster, do the answers to (a) and (b) increase, decrease, or remain the same?

See all solutions

Recommended explanations on Physics 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.