/*! 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 22 One of the fastest recorded pitc... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

One of the fastest recorded pitches in major-league baseball, thrown by Joel Zumaya in 2006, was clocked at \(101.0 \mathrm{mi} / \mathrm{h}\) (Fig. P3.22). If a pitch were thrown horizonLally with this velocity, how far would the ball fall vertically by the time it reached home plate, \(60.5 \mathrm{ft}\) away?

Short Answer

Expert verified
The baseball falls approximately 2.7 feet vertically by the time it reaches home plate.

Step by step solution

01

Convert Units

First, convert the units of speed and distance to be consistent. The distance to home plate is given in feet, and the speed is given in miles per hour. Convert speed into feet per second. For reference, \(1 \mathrm{mi/hr} = 1.467 \mathrm{ft/sec}\). Calculate the speed of the ball in feet per second: \(Speed = 101.0 \mathrm{mi/hr} * 1.467 \mathrm{ft/sec} = 148.267 \mathrm{ft/sec}\).
02

Calculate Time

Use the distance and speed to calculate the time it takes for the ball to reach home plate. The formula is \(d = vt\), rearrange it to solve for time \(t = d/v\). So, \(t = 60.5 \mathrm{ft} / 148.267 \mathrm{ft/sec} = 0.408 \mathrm{sec}\).
03

Calculate Vertical Distance

Use the time obtained in Step 2 to calculate the vertical distance fallen due to gravity. The formula is \(h = 0.5gt^2\), where \(g = 32.174 \mathrm{ft/sec}^2\) is the acceleration due to gravity. So, \(h = 0.5 * 32.174 \mathrm{ft/sec}^2 * (0.408 \mathrm{sec})^2 = 2.7 \mathrm{ft} \). The baseball would fall 2.7 feet vertically while it's traveling to the home plate.

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.

Converting Units in Physics
Understanding the process of converting units is essential in physics and engineering, because it ensures that calculations are accurate and consistent. Let's take a baseball pitch as an example. The pitch is clocked at an impressive speed of 101.0 miles per hour (mi/h), but to solve the problem of how far the ball will fall vertically, we need to work with the speed in feet per second (ft/s).

To convert mph to ft/s, we use a conversion factor. For every mile per hour, there are approximately 1.467 feet per second. Therefore, the conversion is a straightforward multiplication: the speed of the pitch in ft/s is 101.0 mi/h multiplied by 1.467 ft/s per mi/h, which equals 148.267 ft/s. This step is crucial since it aligns the units of speed with the given distance to home plate in feet, allowing us to apply the kinematic equations accurately.
Kinematic Equations
The kinematic equations offer a way to describe the motion of objects using mathematical formulas. They are pivotal in solving problems related to displacement, velocity, acceleration, and time. In the case of projectile motion, which is our current topic, these equations help determine how objects move under the influence of gravity alone.

In the baseball example, we use the simple form of the kinematic equation for velocity and distance, d = vt, where d is the distance, v is the velocity, and t is the time. Rearranging for time, we get t = d/v. With the distance to home plate and the horizontal velocity converted into consistent units, we can find out how long it takes for the baseball pitched horizontally at 148.267 ft/s to travel 60.5 feet. The answer is approximately 0.408 seconds. This information is necessary to determine how far the ball will fall due to gravity over this time interval.
Acceleration Due to Gravity
The acceleration due to gravity, denoted by g, is a constant value that represents the rate at which an object accelerates towards the Earth when it is in free fall. On the surface of the Earth, this acceleration is approximately 32.174 feet per second squared (ft/s²). This fundamental physical constant is a vital component when solving problems related to projectile motion.

To calculate the vertical distance a baseball falls during a pitch, we use the kinematic equation that incorporates the acceleration due to gravity: h = 0.5gt². Since we've already determined that it takes 0.408 seconds for the ball to reach home plate horizontally, we can calculate the vertical distance fallen in that time: h = 0.5 * 32.174 ft/s² * (0.408 sec)², resulting in a vertical drop of approximately 2.7 feet. This calculation shows that even over a relatively short horizontal distance, gravity significantly affects the vertical position of a projectile.

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

One strategy in a snowball fight is to throw a snowball at a high angle over level ground. Then, while your opponent is watching that snowball, you throw a second one at a low angle timed to arrive before or at the same time as the first one. Assume both snowballs are thrown with a speed of \(25.0 \mathrm{~m} / \mathrm{s}\). The first is thrown at an angle of \(70.0^{\circ}\) with respect to the horizontal. (a) At what angle should the second snowball be thrown to arrive at the same point as the first? (b) How many seconds later should the second snowball be thrown after the first in order for both to arrive at the same time?

A quarterback throws a football toward a receiver with an initial speed of \(20 \mathrm{~m} / \mathrm{s}\) at an angle of \(30^{\circ}\) above the horizontal. At that instant the receiver is \(20 \mathrm{~m}\) from the quarterback. In what direction and with what constant speed should the receiver run in order to catch the football at the level at which it was thrown?

A jet airliner moving initially at \(3.00 \times 10^{2} \mathrm{mi} / \mathrm{h}\) due east enters a region where the wind is blowing \(1.00\) \(\times 10^{2} \mathrm{mi} / \mathrm{h}\) in a direction \(30.0^{\circ}\) north of east. (a) Find the components of the velocity of the jet airliner relative to the air, \(\vec{v}_{\mu}\). (b) Find the components of the velocity of the air relative to Earth, \(\overrightarrow{\mathbf{v}}_{M \mathcal{E}}\). (c) Write an equation analo- gous to Equation \(3.16\) for the velocities \(\vec{v}_{j}, \vec{v}_{A B}\), and \(\vec{v}_{g E}\) (d) What is the speed and direction of the aircraft relative to the ground?

A rocket is launched at an angle of \(53.0^{\circ}\) above the horizontal with an initial speed of \(100 \mathrm{~m} / \mathrm{s}\). The rocket moves for \(3.00 \mathrm{~s}\) along its initial line of motion with an acceleration of \(30.0 \mathrm{~m} / \mathrm{s}^{2}\). At this time, its engines fail and the rocket proceeds to move as a projectile. Find (a) the maximum altitude reached by the rocket, (b) its total time of flight, and \((c)\) its horizontal range.

The magnitude of vector \(\vec{A}\) is \(35.0\) units and points in the direction \(325^{\circ}\) counterclockwise from the positive \(x\) -axis. Calculate the \(x\) -and \(y\) -components of this vector.

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.