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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.
This formula highlights a key relationship between the angle of launch \( \theta \), the initial velocity, and the effect of gravity on the projectile's path.The effectiveness of this formula is in its ability to provide insights into the physics of motion. By altering the angle or speed, one can predict the resulting change in range. This equation captures how crucial the interplay between angle and velocity is to the projectile's flight.
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:
  • Finding the angle \( \theta \) that results in \( \sin(2\theta) = 1 \).
  • This occurs at \( 2\theta = 90^\circ \) resulting in \( \theta = 45^\circ \).
By understanding that lifting the launch angle to 45 degrees achieves this, we optimize for the greatest possible range, assuming constant initial speed and gravitational force. It's essential to remember, however, that in real-world conditions, other factors such as air resistance might alter the optimal angle slightly.
This theoretical analysis is foundational for interpreting various real-world scenarios, from sports to engineering, where understanding projectile range is crucial.

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Most popular questions from this chapter

A soccer ball is kicked from the ground with an initial speed of \(12 \mathrm{~m} / \mathrm{s}\) at an angle of \(32^{\circ}\) above the horizontal. What are the \(x\) and \(y\) positions of the ball \(0.50 \mathrm{~s}\) after it is kicked?

For each of the following quantities, indicate whether it is a scalar or a vector: (a) the time it takes you to run the \(100-m\) dash, (b) your displacement after running the \(100-\mathrm{m}\) dash, (c) your average velocity while running, (d) your average speed while running.

Vector \(\overrightarrow{\mathbf{A}}\) points in the negative \(y\) direction and has a magnitude of \(5 \mathrm{~km}\). Vector \(\overrightarrow{\mathbf{B}}\) has a magnitude of \(15 \mathrm{~km}\) and points in the positive \(x\) direction. Use components to find the magnitude of (a) \(\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}\), (b) \(\overrightarrow{\mathbf{A}}-\overrightarrow{\mathbf{B}}\), and (c) \(\overrightarrow{\mathbf{B}}-\overrightarrow{\mathbf{A}}\).

The press box at a baseball park is \(9.75 \mathrm{~m}\) above the ground. A reporter in the press box looks at an angle of \(15.0^{\circ}\) below the horizontal to see second base. What is the horizontal distance from the press box to second base?

The initial velocity of a projectile has a horizontal component equal to \(5 \mathrm{~m} / \mathrm{s}\) and a vertical component equal to \(6 \mathrm{~m} / \mathrm{s}\). At the highest point of the projectile's flight, what is (a) the horizontal component of its velocity and (b) the vertical component of its velocity? Explain.

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