/*! 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 51 A batter hits a baseball at spee... [FREE SOLUTION] | 91Ó°ÊÓ

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A batter hits a baseball at speed \(v\), at an angle \(\theta\) above horizontal. (a) Find an equation for the range (horizontal distance to where the ball falls), \(R\), in terms of the relevant variables. Neglect air friction and the height of the ball above the ground when it is hit. \hwans\\{hwans:baseballrange\\} (b) Interpret your equation in the cases of \(\theta=0\) and \(\theta=90^{\circ}\). (c) Find the angle that gives the maximum range. Wwans\\{hwans:baseballrange\\}

Short Answer

Expert verified
The range equation is \(R = \frac{v^2 \sin 2\theta}{g}\). At \(\theta=0^{\circ}\) and \(\theta=90^{\circ}\), the range is 0. The maximum range occurs at \(\theta=45^{\circ}\).

Step by step solution

01

Understand the problem

We are tasked with finding the horizontal range of a baseball hit at a speed \(v\) and an angle \(\theta\). Neglect factors like air resistance and initial height.
02

Identify the relevant equations

The range \(R\) of a projectile can be found using the equation: \[ R = \frac{v^2 \sin 2\theta}{g} \]where \(g\) is the acceleration due to gravity (approximately \(9.8 \, \text{m/s}^2\)).
03

Derive the range equation

Using the known projectile motion formula, start by resolving the initial velocity \(v\) into its components: - Horizontal velocity: \(v_x = v \cos\theta\)- Vertical velocity: \(v_y = v \sin\theta\).The time \(t\) for the flight is given by \(t = \frac{2v_y}{g} = \frac{2v \sin\theta}{g}\).Using horizontal motion, the range \(R\) is \(v_x \times t = v \cos\theta \times \frac{2v \sin\theta}{g} = \frac{2v^2 \sin\theta \cos\theta}{g}\).Utilize the trigonometric identity \(2 \sin\theta \cos\theta = \sin 2\theta\); substitute to get \(R = \frac{v^2 \sin 2\theta}{g}\).
04

Interpret the cases of \(\theta=0^{\circ}\) and \(\theta=90^{\circ}\)

For \(\theta=0^{\circ}\), the range \(R\) becomes \(R = \frac{v^2 \sin 0}{g} = 0\).For \(\theta=90^{\circ}\), the range \(R\) becomes \(R = \frac{v^2 \sin 180}{g} = 0\).In both scenarios, the ball either doesn’t go up or comes straight down, resulting in zero horizontal range.
05

Find the angle for maximum range

The range \(R = \frac{v^2 \sin 2\theta}{g}\) is maximized when \(\sin 2\theta = 1\) since the maximum value for the sine function is 1.This occurs when \(2\theta = 90^{\circ}\), or \(\theta = 45^{\circ}\).

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Key Concepts

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

Projectile Range
Projectile range is the horizontal distance a projectile travels before it hits the ground. It is a crucial concept in understanding how far an object can travel under ideal conditions, free from air resistance or other forces. For a baseball hit with a speed \(v\) and at an angle \(\theta\) to the horizontal, the range \(R\) is described by the equation:\[ R = \frac{v^2 \sin 2\theta}{g} \]where \(g\) is the acceleration due to gravity (approximately \(9.8\, \text{m/s}^2\)).To determine the projectile range:
  • Identify the initial speed \(v\).
  • Determine the launch angle \(\theta\).
  • Use the above formula to calculate the range.
In scenarios where \(\theta = 0^\circ\) or \(\theta = 90^\circ\), the range becomes zero because the projectile either doesn’t leave the ground or moves straight up and then back down.
Trigonometric Identities
Trigonometric identities play a key role in simplifying projectile range formulas. In our case, the identity \(\sin 2\theta = 2 \sin\theta \cos\theta\) is particularly useful.When calculating projectile range, we often break down the initial velocity into components:
  • Horizontal velocity: \(v_x = v \cos\theta\)
  • Vertical velocity: \(v_y = v \sin\theta\)
Using trigonometric identities helps in transforming the product of sine and cosine into a single sine function of a double angle, simplifying the otherwise complex multiplication of these terms.By substituting \(\sin 2\theta\), we can derive the range equation more efficiently in terms of fewer variables and trigonometric functions.
Optimal Launch Angle
Finding the optimal launch angle means determining the angle \(\theta\) that maximizes the projectile's range. The formula for range \(R = \frac{v^2 \sin 2\theta}{g}\) highlights that the range depends heavily on the \(\sin 2\theta\).The sine function reaches its maximum value of 1 at an angle of \(90^\circ\). Thus, for the maximum range:
  • Set \(2\theta = 90^\circ\), which means \(\theta = 45^\circ\).
At \(\theta = 45^\circ\), we achieve the longest horizontal distance possible for given initial speed \(v\), as it balances the initial velocity between vertical and horizontal components effectively.
Kinematic Equations
Kinematic equations describe the motion of objects, accounting for variables like velocity, time, and distance. In projectile motion, they help us predict the trajectory and maximum range of the projectile.To understand these, consider the following components of motion:
  • Horizontal Motion: The horizontal velocity occurs at a constant speed, calculated by \(v_x = v \cos\theta\).
  • Vertical Motion: Accelerated by gravity. The vertical velocity is \(v_y = v \sin\theta\), and the time of flight can be calculated by \(t = \frac{2v_y}{g}\).
Using these components, we integrate both types of motion into the projectile range formula, thus providing an equation that predicts the projectile distance based on launch conditions.These kinematic principles are fundamental in ensuring that our approach to calculating projectile range accurately represents real-world physics.

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Most popular questions from this chapter

A blimp is initially at rest, hovering, when at \(t=0\) the pilot turns on the motor of the propeller. The motor cannot instantly get the propeller going, but the propeller speeds up steadily. The steadily increasing force between the air and the propeller is given by the equation \(F=k t\), where \(k\) is a constant. If the mass of the blimp is \(m\), find its position as a function of time. (Assume that during the period of time you're dealing with, the blimp is not yet moving fast enough to cause a significant backward force due to air resistance.)(answer check available at lightandmatter.com)

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At low speeds, every car's acceleration is limited by traction, not by the engine's power. Suppose that at low speeds, a certain car is normally capable of an acceleration of \(3 \mathrm{~m} / \mathrm{s}^{2}\). If it is towing a trailer with half as much mass as the car itself, what acceleration can it achieve? [Based on a problem from PSSC Physics.]

(solution in the pdf version of the book) A car is accelerating forward along a straight road. If the force of the road on the car's wheels, pushing it forward, is a constant \(3.0 \mathrm{kN}\), and the car's mass is \(1000 \mathrm{~kg}\), then how long will the car take to go from \(20 \mathrm{~m} / \mathrm{s}\) to \(50 \mathrm{~m} / \mathrm{s} ?\)

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