Question

Projectile Distance An object is fired at an angle \(\theta\) to the horizontal with an initial speed of \(v_{0}\) feet per second. Ignoring air resistance, the length of the projectile's path is given by $$ L(\theta)=\frac{v_{0}^{2}}{32}\left\\{\sin \theta-\cos ^{2} \theta\left(\ln \left[\tan \left(\frac{\pi-2 \theta}{4}\right)\right]\right)\right\\} $$ where \(0<\theta<\frac{\pi}{2}\) (a) Find the length of the object's path for angles \(\theta=\frac{\pi}{6}, \frac{\pi}{4}\), and \(\frac{\pi}{3}\) if the initial velocity is 128 feet per second. (b) Using a graphing utility, determine the angle required for the object to have a path length of 550 feet if the initial velocity is 128 feet per second. (c) What angle will result in the longest path? How does this angle compare to the angle that results in the longest range? (See Problems \(63-66\) in Section \(7.3 .)\)

Short answer

To find the length for each angle, substitute the value of \( \theta \) into the equation and simplify it. Use a graphing utility to find the angle that gives a path length of 550 feet and to determine the angle for the longest path.

Step-by-step solution

Given Information

The length of the projectile's path is given by the equation:\[ L(\theta) = \frac{v_{0}^{2}}{32} \bigg( \text{sin} \theta - \text{cos}^{2} \theta \big( \text{ln} \big[ \text{tan} \big( \frac{\text{π} - 2\theta}{4} \big) \big] \big) \bigg) \]and the initial velocity is \( v_{0} = 128 \) feet per second. We are given angles \( \theta = \frac{\text{π}}{6} \), \( \frac{\text{π}}{4} \), and \( \frac{\text{π}}{3} \).

Calculate the Length for \( \theta = \frac{\text{π}}{6} \)

Substitute \( \theta = \frac{\text{π}}{6} \) into the equation:\[ L\big( \frac{\text{π}}{6} \big) = \frac{128^2}{32} \bigg( \text{sin} \big( \frac{\text{π}}{6} \big) - \text{cos}^{2} \big( \frac{\text{π}}{6} \big) \big( \text{ln} \big[ \text{tan} \big( \frac{\text{π} - 2 \times \frac{\text{π}}{6}}{4} \big) \big] \big) \bigg) \]Given that \( \text{sin} \big( \frac{\text{π}}{6} \big) =\frac{1}{2} \) and \( \text{cos} \big( \frac{\text{π}}{6} \big) = \frac{\text{√3}}{2} \), we can simplify and calculate the length.

Calculate the Length for \( \theta = \frac{π}{4} \)

Substitute \( \theta = \frac{\text{π}}{4} \) into the equation:\[ L\big( \frac{\text{π}}{4} \big) = \frac{128^2}{32} \bigg( \text{sin} \big( \frac{\text{π}}{4} \big) - \text{cos}^{2} \big( \frac{\text{π}}{4} \big) \big( \text{ln} \big[ \text{tan} \big( \frac{\text{π} - 2 \times \frac{\text{π}}{4}}{4} \big) \big] \big) \bigg) \]Given that \( \text{sin} \big( \frac{\text{π}}{4} \big) = \frac{\text{√2}}{2} \) and \( \text{cos} \big( \frac{\text{π}}{4} \big) = \frac{\text{√2}}{2} \), we can simplify and calculate the length.

Calculate the Length for \( \theta = \frac{π}{3} \)

Substitute \( \theta = \frac{\text{π}}{3} \) into the equation:\[ L\big( \frac{\text{π}}{3} \big) = \frac{128^2}{32} \bigg( \text{sin} \big( \frac{\text{π}}{3} \big) - \text{cos}^{2} \big( \frac{\text{π}}{3} \big) \big( \text{ln} \big[ \text{tan} \big( \frac{\text{π}}{3} \big) \big] \big) \bigg) \]Given that \( \text{sin} \big( \frac{\text{π}}{3} \big) = \frac{\text{√3}}{2} \) and \( \text{cos} \big( \frac{\text{π}}{3} \big) = \frac{1}{2} \), we can simplify and calculate the length.

Using a Graphing Utility

Use a graphing utility to plot \( L(\theta) \) as a function of \( \theta \) and find the angle \( \theta \) required for the path length to be 550 feet when the initial velocity is 128 feet per second.

Determine the Longest Path Angle

Using the graph from the previous step, find the angle \( \theta \) that maximizes the path length. Compare this angle to the angle that results in the longest range, which is typically \( \theta = \frac{\text{π}}{4} \) for projectile motion in the absence of air resistance.

Key concepts

Trigonometric Functions

Understanding trigonometric functions is essential for solving problems involving projectile motion. Trigonometric functions like sine (\text{sin}) and cosine (\text{cos}) relate the angles of a triangle to the lengths of its sides. In the context of projectile motion:

• Sine (\text{sin}) of an angle \(\theta\) refers to the ratio of the length of the opposite side to the hypotenuse in a right triangle.
• Cosine (\text{cos}) of an angle \(\theta\) refers to the ratio of the length of the adjacent side to the hypotenuse.

The values of these functions for commonly used angles such as \( \frac{\pi}{6} \), \( \frac{\pi}{4} \), and \( \frac{\pi}{3} \) are often memorized:

• \( \text{sin} \left( \frac {\pi}{6} \right) = \frac{1}{2} \)
• \( \text{sin} \left( \frac {\pi}{4} \right) = \frac{ \sqrt {2}}{ 2} \)
• \( \text{sin} \left( \frac {\pi}{3} \right) = \frac{ \sqrt{3}}{2} \)
• \( \text{cos} \left( \frac {\pi}{6} \right) = \frac{ \sqrt {3}}{ 2} \)
• \( \text{cos} \left( \frac {\pi}{4} \right) = \frac{ \sqrt {2}}{ 2} \)
• \( \text{cos} \left( \frac {\pi}{3} \right) = \frac{1}{2} \)

These functions help in substituting the angles into the given formula for projectile path length to find the desired values for different angles.

Angle Optimization

Optimizing the angle \(\theta\) is crucial for achieving different objectives in projectile motion, such as maximum range or a specific path length. Here are some key considerations:

• Maximum Range: For projectile motion with no air resistance, the angle that maximizes the range (horizontal distance) is typically \( \theta = \frac {\pi}{4} \) or 45 degrees. This is because the initial velocity's components are equal in both the horizontal and vertical directions, providing optimal balance.
• Specific Path Length: When aiming for a particular path length such as 550 feet, it's important to use a graphing utility to plot the path length function,\( L( \theta ) \), and find the corresponding angle for the given initial velocity. This graphical approach allows you to visually determine the required angle by identifying where the curve intersects the desired length.

In the given problem, using a graphing calculator or software helps determine the exact angle needed for the path length of 550 feet when the initial velocity is 128 feet per second.

Initial Velocity

Initial velocity \( v_{0} \) significantly influences the projectile's path length and behavior. Here are key aspects to understand:
* **Definition:** Initial velocity is the speed at which the object is launched. It affects both the horizontal and vertical components of motion.
.

• Vertical Component: This component determines how high the projectile will rise. It is calculated as
  \[ v_{0y} = v_{0} \sin ( \theta ). \]
• Horizontal Component: This component determines how far the projectile will travel horizontally. It is given by
  \[ v_{0x} = v_{0} \cos ( \theta ). \]

* **In the given problem:** The initial velocity is specified as 128 feet per second. By substituting this value into the path length formula together with the corresponding trigonometric functions for given angles, one can calculate the projectile path length for specific launch angles. For different initial velocities, the path length and range change accordingly while keeping other aspects, such as angle and acceleration due to gravity, constant.