Problem 59 - Contemporary Precalculus

Question

Use a calculator in degree mode and assume that air resistance is negligible. A medieval bowman shoots an arrow which leaves the bow 4 feet above the ground with an initial velocity of 88 feet per second at an angle of \(48^{\circ}\) with the horizontal. (a) Graph the arrow's path. (b) Will the arrow go over the 40 -foot-high castle wall that is 200 feet from the archer?

Step-by-Step Solution

Answer

Answer: ____________ (Provide a brief explanation based on the solution steps above)

Step 1: Determine initial velocities

We know the initial speed of the arrow is 88 ft/s, and the launch angle is \(48^{\circ}\). We will use the following trigonometric formulas to find the initial horizontal and vertical velocities: - \(v_{0x} = v_{0}\text{cos}\theta\) - \(v_{0y} = v_{0}\text{sin}\theta\) In this case, \(v_{0} = 88 ft/s\), \(\theta=48^{\circ}\).

Step 2: Calculate the time to reach the ground

The arrow's vertical position can be determined using the equation \(y(t) = y_{0} + v_{0y}t - \frac{1}{2}gt^{2}\), where \(y_{0}=4 ft\) is the initial height, \(v_{0y}\) is the initial vertical velocity, \(g=32 ft/s^{2}\) is the acceleration due to gravity, and \(t\) is the time. Since the arrow reaches the ground when \(y(t)=0\), we need to find \(t\) that satisfies this equation.

Step 3: Find the horizontal position

The horizontal position of the arrow can be determined using the equation \(x(t) = x_{0} + v_{0x}t\), where \(x_{0}=0\) is the initial position and \(v_{0x}\) is the initial horizontal velocity. Plotting \(y(t)\) against \(x(t)\), we can draw the arrow's path. (b) Will the arrow go over the 40-ft-high castle wall that is 200 feet from the archer?

Step 4: Calculate the time when the arrow reaches the wall

Since the castle wall is 200 feet from the archer, we will use the equation \(x(t) = x_{0} + v_{0x}t\) to determine the time when the arrow reaches that position. Plugging in the value of \(x=200 ft\), \(x_{0}=0\), and \(v_{0x}\), we find the corresponding time.

Step 5: Calculate the height at that time

Now, we will use the same time found in Step 4 to determine the height of the arrow using the equation \(y(t) = y_{0} + v_{0y}t - \frac{1}{2}gt^{2}\). If the height(\(y(t)\)) is greater than the height of the wall (40 ft), then the arrow will go over the castle wall; otherwise, it will not.

Key Concepts

Trigonometry

Understanding trigonometry is essential when analyzing projectile motion. In this exercise, the angle of launch is a key factor in determining the initial horizontal and vertical components of velocity. Trigonometry lets us break down any vector, such as the arrow's velocity, into horizontal and vertical parts.

Here's how it works:

• For the horizontal velocity (\(v_{0x}\)), we use the cosine of the angle: \(v_{0x} = v_{0} \cos \theta\). The cosine function helps us understand how much of the total velocity is directed horizontally.
• For the vertical velocity (\(v_{0y}\)), we employ the sine of the angle: \(v_{0y} = v_{0} \sin \theta\). The sine function describes the portion of the velocity that pushes the arrow upwards.

In this specific problem, the launch angle is \(48^{\circ}\), allowing us to accurately find both components of velocity.
By knowing these, we can predict how far and how high the arrow will travel.

Initial Velocity

Initial velocity plays a major role in projectile motion. It is the speed at which the arrow leaves the bow. In this task, the initial velocity is given as 88 feet per second. This value determines the starting kinetic energy and influences both how high and how far the projectile travels.

With initial velocity, we calculate both horizontal and vertical components:

• Initial horizontal velocity (\(v_{0x}\)) is responsible for moving the arrow forward.
• Initial vertical velocity (\(v_{0y}\)) dictates how high the arrow will travel.

These components are combined with the angle of launch to define the arrow's flight path.
In equations involving projectile motion, the initial velocity is often denoted as \(v_{0}\). Modifying either the magnitude or angle would significantly change the trajectory, demonstrating its crucial influence.

Trajectory Calculation

The trajectory of a projectile is the path it follows when it is launched into the air. This path can be calculated using mathematical formulas that involve both time and the velocities we have previously determined.

For trajectory calculations, we use:

• The horizontal position formula: \(x(t) = x_{0} + v_{0x} t\). Here, \(x_{0}\) represents the initial position (usually \(0\) unless stated otherwise).
• The vertical position formula: \(y(t) = y_{0} + v_{0y} t - \frac{1}{2} g t^{2}\). The term \(y_{0}\) denotes the initial vertical position (in this problem, 4 feet above the ground), while \(g\) is the acceleration due to gravity (32 feet per second squared).

Graphing the results of these calculations helps visualize the arrow’s path.
By analyzing these values at specific times, we determine whether the arrow will encounter obstacles, such as a castle wall, in its path.

Mathematical Modeling

Mathematical modeling allows us to predict real-world scenarios using equations. In this example, mathematical modeling helps us visualize whether an arrow can clear a castle wall.

This involves several steps:

• Modeling the problem with physics equations for motion in two dimensions, considering factors like gravity and initial conditions.
• Using variables to describe initial conditions (like velocity and height) and natural forces (like gravity).
• Solving these equations to predict outcomes, such as whether or not the arrow will pass a 40-foot-high wall.

Through mathematical modeling, we convert a physical problem into a solvable mathematical form. This approach is widely used in physics and engineering for trajectory and motion analysis, making it an invaluable tool for any student.