Chapter 4: Problem 44
How must a projectile be launched in order to maximize its range?
Short Answer
Expert verified
Launch the projectile at a 45-degree angle to maximize its range.
Step by step solution
01
Identify the Range Formula
The range of a projectile, launched from the ground with initial speed \( v_0 \) and angle \( \theta \) with respect to the horizontal, is given by the formula: \[ R = \frac{v_0^2 \sin(2\theta)}{g} \]where \( g \) is the acceleration due to gravity. We need to maximize the range \( R \).
02
Find the Derivative
To maximize the range, we must find the maximum value of \( \sin(2\theta) \). Consider that the derivative of \( \sin(2\theta) \) with respect to \( \theta \) is given by:\[ \frac{d}{d\theta} \sin(2\theta) = 2 \cos(2\theta) \] Set this to zero to find critical points.
03
Solve for Critical Points
Set the derivative equal to zero and solve for \( \theta \):\[ 2 \cos(2\theta) = 0 \]This implies:\[ \cos(2\theta) = 0 \] The solutions to this are \( 2\theta = \frac{\pi}{2} + n\pi \) for integer values of \( n \).
04
Simplify to Find Optimal Angle
To simplify, use \( n = 0 \):\[ 2\theta = \frac{\pi}{2} \] \[ \theta = \frac{\pi}{4} \] Thus, for maximizing the range, the projectile should be launched at an angle of \( \frac{\pi}{4} \) or 45 degrees.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Range Formula
When studying projectile motion, understanding the range formula is crucial. The formula allows us to calculate how far a projectile will travel along the horizontal plane. It is given by:\[R = \frac{v_0^2 \sin(2\theta)}{g}\]where:
- \( R \) is the range or horizontal distance traveled by the projectile.
- \( v_0 \) is the initial speed of the projectile.
- \( \theta \) is the launch angle measured from the horizontal.
- \( g \) is the acceleration due to gravity, approximately \( 9.8 \text{ m/s}^2 \) on Earth.
Optimal Angle
Finding the best launch angle for maximum range is an interesting challenge in projectile motion. From the range formula, we look to maximize the term \( \sin(2\theta) \). The sine function reaches its maximum value of 1 when \( 2\theta = \frac{\pi}{2} \) radians, or 90 degrees. This indicates that \( \theta = \frac{\pi}{4} \) or 45 degrees is the optimal angle.The process involves basic calculus where we set the derivative of \( \sin(2\theta) \) with respect to \( \theta \) to zero:\[\frac{d}{d\theta} \sin(2\theta) = 2 \cos(2\theta) = 0\]Solving this equation confirms the critical points of the sine function, leading us to the conclusion that a 45-degree angle maximizes the range.This angle is considered optimal because, at 45 degrees, the projectile splits its initial speed between the vertical and horizontal components evenly. This balance ensures the longest horizontal distance covered before the projectile hits the ground.
Maximizing Range
Maximizing range is a fundamental concern when dealing with projectile motion problems. Based on the range formula \( R = \frac{v_0^2 \sin(2\theta)}{g} \), the key to maximizing \( R \) lies in making \( \sin(2\theta) \) as large as possible.
In essence, the largest value the sine function can take is 1. This means the challenge becomes:
This theoretical analysis is foundational for interpreting various real-world scenarios, from sports to engineering, where understanding projectile range is crucial.
In essence, the largest value the sine function can take is 1. This means the challenge becomes:
- Finding the angle \( \theta \) that results in \( \sin(2\theta) = 1 \).
- This occurs at \( 2\theta = 90^\circ \) resulting in \( \theta = 45^\circ \).
This theoretical analysis is foundational for interpreting various real-world scenarios, from sports to engineering, where understanding projectile range is crucial.