/*! 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 3 A baseball player hits a 5.1 -oz... [FREE SOLUTION] | 91Ó°ÊÓ

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A baseball player hits a 5.1 -oz baseball with an initial velocity of \(140 \mathrm{ft} / \mathrm{s}\) at an angle of \(40^{\circ}\) with the horizontal as shown. Determine \((a)\) the kinetic energy of the ball immediately after it is hit, \((b)\) the kinetic energy of the ball when it reaches its maximum height, maximum height above the ground reached by the ball.

Short Answer

Expert verified
(a) 97.02 ft·lb, (b) 57.05 ft·lb, maximum height: 125.24 ft.

Step by step solution

01

Convert weight to mass

The weight of the baseball is given as 5.1 ounces. First, convert the weight to mass in pounds because weight is a force. Use the conversion factor: 1 oz = 1/16 lb. Thus, \( \text{mass} = \frac{5.1}{16} \text{ lb} \approx 0.31875 \text{ lb} \). Also, using gravitational acceleration \( g = 32.2 \text{ ft/s}^2 \), convert the weight to mass in slugs: \( \text{mass} = \frac{0.31875 \text{ lb}}{32.2 \text{ ft/s}^2} \approx 0.0099 \text{ slugs} \).
02

Calculate initial kinetic energy

The kinetic energy \( KE \) is given by the formula \( KE = \frac{1}{2}mv^2 \). Here, \( m = 0.0099 \text{ slugs} \) and \( v = 140 \text{ ft/s} \). Calculate: \(KE = \frac{1}{2} \times 0.0099 \times (140)^2 = 97.02 \text{ ft} \cdot \text{lb} \).
03

Determine vertical component of velocity

Find the vertical component of the initial velocity using \( v_{y} = v \cdot \sin(\theta) \). Substitute \( v = 140 \text{ ft/s} \) and \( \theta = 40^{\circ} \):\( v_{y} = 140 \cdot \sin(40^{\circ}) \approx 89.89 \text{ ft/s} \).
04

Calculate kinetic energy at maximum height

At maximum height, the vertical component of the velocity is 0. The horizontal component remains unchanged, given by \( v_{x} = v \cdot \cos(\theta) \). Hence, \( v_{x} = 140 \cdot \cos(40^{\circ}) \approx 107.23 \text{ ft/s} \). Therefore, the kinetic energy at maximum height is:\( KE_{max} = \frac{1}{2} \times 0.0099 \times (107.23)^2 \approx 57.05 \text{ ft} \cdot \text{lb} \).
05

Calculate maximum height

Use the kinematic equation to find the maximum height: \( v_y^2 = v_{y0}^2 - 2g h \), where \( v_y = 0 \) at the maximum height. Solve for \( h \):\( 0 = (89.89)^2 - 2 \times 32.2 \times h \). Thus, \( h = \frac{(89.89)^2}{2 \times 32.2} \approx 125.24 \text{ ft} \).

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Key Concepts

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

Kinematics
Kinematics is the branch of physics that deals with the motion of objects without considering the forces that cause this motion. It focuses on parameters like velocity, acceleration, displacement, and time.
In our exercise, we analyze the motion of a baseball after it is hit. We know the ball's initial velocity and angle of projection. This information allows us to decompose its velocity into horizontal and vertical components, which describe how the motion unfolds.
Utilizing kinematic equations helps solve for various parameters, such as the maximum height reached by the projectile or its range. Understanding these basic principles can assist in solving problems regarding any object in motion.
Projectile Motion
Projectile motion refers to the motion of an object thrown or projected into the air, subject to only the acceleration of gravity. This type of motion follows a parabolic path, due to the constant gravitational force acting on the object.
When the baseball player hits the ball, it follows a path determined by its initial velocity and angle. These factors allow us to determine the projectile's vertical and horizontal components using trigonometric functions:
  • Vertical component: \( v_{y} = v \cdot \sin(\theta) \)
  • Horizontal component: \( v_{x} = v \cdot \cos(\theta) \)
Using these components, we can calculate the motion aspects, such as maximum height and horizontal distance traveled (range). Understanding these principles is crucial for solving any projectile motion problem.
Energy Conservation
The principle of energy conservation states that the total energy in a closed system remains constant. In other words, energy cannot be created or destroyed; it can only be transformed from one form to another. This concept is vital in solving dynamics problems involving movement.
For the baseball in motion, the kinetic energy initially consists of both the horizontal and vertical components of velocity. As the baseball ascends to its peak height, all the vertical kinetic energy transforms into gravitational potential energy. At maximum height, the vertical component ceases, yet the horizontal component and corresponding kinetic energy remain constant.
This understanding permits precise calculations of the kinetic energy at any point in its trajectory, particularly when it reaches maximum height.
Conversion of Units
Converting units is necessary when dealing with physics problems to ensure all measurements are consistent. Often, measurements are given in different units, and it’s essential to unify these units either into the International System of Units (SI) or another desired system.
In this exercise, the weight of the baseball is initially given in ounces. Subsequently, we convert it to pounds and finally to slugs to align with the equations involving mass and velocity. This adjustment allows us to accurately calculate the kinetic energy:
  • 1 ounce = 1/16 pounds
  • 1 pound = 0.031 slugs (using gravitational acceleration \( g = 32.2 \text{ ft/s}^2 \))
Effective conversion of units ensures the calculations adhere to the same measurement system, leading to accurate results.

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Most popular questions from this chapter

Packages in an automobile parts supply house are transported to the loading dock by pushing them along on a roller track with very little friction. At the instant shown, packages B and C are at rest, and package A has a velocity of 2 m/s. Knowing that the coefficient of restitution between the packages is 0.3, determine (a) the velocity of package C after A hits B and B hits C, (b) the velocity of A after it hits B for the second time.

A ballistic pendulum is used to measure the speed of high-speed projectiles. A \(6-\mathrm{g}\) bullet \(A\) is fired into a \(1-\mathrm{kg}\) wood block \(B\) suspended by a cord with a length of \(l=2.2 \mathrm{m}\). The block then swings through a maximum angle of \(\theta=60^{\circ} .\) Determine \((a)\) the initial speed of the bullet \(v_{0},(b)\) the impulse imparted by the bullet on the block, \((c)\) the force on the cord immediately after the impact.

A 2-kg sphere moving to the right with a velocity of 5 m/s strikes at A, which is on the surface of a 9-kg quarter cylinder that is initially at rest and in contact with a spring with a constant of 20 kN/m. The spring is held by cables, so it is initially compressed 50 mm. Neglecting friction and knowing that the coefficient of restitution is 0.6, determine (a) the velocity of the sphere immediately after impact, (b) the maximum compressive force in the spring.

Ball \(B\) is hanging from an inextensible cord. An identical ball \(A\) is released from rest when it is just touching the cord and drops through the vertical distance \(h_{A}=8\) in. before striking ball \(B .\) Assuming \(e=0.9\) and no friction, determine the resulting maximum vertical displacement \(h_{B}\) of the ball \(B .\)

A 25 -ton railroad car moving at \(2.5 \mathrm{mi} / \mathrm{h}\) is to be coupled to a 50 -ton car that is at rest with locked wheels \(\left(\mu_{k}=0.30\right) .\) Determine \((a)\) the velocity of both cars after the coupling is completed, (b) the time it takes for both cars to come to rest.

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