Problem 43 - College Trigonometry

Question

Carl's friend Jason competes in Highland Games Competitions across the country. In one event, the 'hammer throw', he throws a 56 pound weight for distance. If the weight is released 6 feet above the ground at an angle of \(42^{\circ}\) with respect to the horizontal with an initial speed of 33 feet per second, find the parametric equations for the flight of the hammer. (Here, use \(\left.g=32 \frac{f t}{s^{2}} .\right)\) When will the hammer hit the ground? How far away will it hit the ground? Check your answer using a graphing utility.

Step-by-Step Solution

Answer

The hammer hits the ground at approximately \(t = 2.26\) seconds and \(x = 55.19\) feet away.

Step 1: Define the Problem

To find the trajectory of the hammer, we'll set up parametric equations for its path, where the horizontal position is described by one equation and the vertical position by another. We will later solve these to find when and where the hammer lands.

Step 2: Establish Initial Conditions

The hammer is released from a height of 6 feet, at an angle of \(42^{\circ}\), and with an initial speed of 33 feet per second. The gravity constant \(g\) is given as \(32 \frac{ft}{s^{2}}\).

Step 3: Construct Parametric Equations

The horizontal distance equation is \(x(t) = (v_0 \cdot \cos(\theta)) t\), where \( v_0 = 33 \) feet per second is the initial speed and \( \theta = 42^{\circ} \). The vertical position equation is \(y(t) = h + (v_0 \cdot \sin(\theta)) t - \frac{1}{2} g t^2\), where \( h = 6 \) is the initial height of the hammer.

Step 4: Calculate the Parameters

Calculate the components:\(v_0 \cdot \cos(\theta) = 33 \cdot \cos(42^\circ)\) and \(v_0 \cdot \sin(\theta) = 33 \cdot \sin(42^\circ)\).

Step 5: Solve for Time of Flight

Set \(y(t) = 0\) to find when the hammer hits the ground: \[6 + (33 \sin(42^\circ)) t - 16t^2 = 0\]. Use the quadratic formula to solve \(8t^2 - 33 \sin(42^\circ)t - 6 = 0\).

Step 6: Evaluate and Simplify

Use the quadratic formula: \[t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]. Here \(a = 8, b = 33 \sin(42^\circ), c = -6\). Calculate the discriminant and solve for \(t\).

Step 7: Calculate Horizontal Distance

Substitute the time \(t\), found when the hammer hits the ground, back into the horizontal position equation \(x(t) = (33 \cos(42^\circ)) t\) to find the horizontal distance.

Step 8: Verify Solutions with a Graphing Utility

Using graphing software, simulate the path described by the parametric equations to verify the time and distance for when the hammer hits the ground.

Key Concepts

Projectile Motion

Projectile motion describes the curved path that an object follows as it moves through the air under the influence of gravity. When you throw a hammer in sports like the Highland Games, or even a ball for that matter, the path it takes is a classic example of projectile motion.

In this exercise, the hammer is thrown at an angle which creates both a horizontal and vertical component to the motion. These two components combine to form the hammer's trajectory. The crucial factors influencing the motion are the angle at which the object is launched (angle of projection), the speed at which it is thrown (initial velocity), and the acceleration due to gravity. Together, these parameters help determine the distance and time it takes for the object to land back on the ground. Understanding these concepts provides a mathematical and visual picture of the projectile's flight.

In summary, projectile motion involves interactions of:

• Initial angle and speed
• Gravity pulling the object downward
• Horizontal distance traveled until it lands

Angle of Projection

The angle of projection is a key factor in determining how far a projectile travels. In our example, the hammer is released at an angle of \(42^{\circ}\) with respect to the horizontal. This angle is essential because it affects the vertical and horizontal distances covered by the projectile.

An angle closer to \(45^{\circ}\) is often considered ideal for maximum distance in projectile motion because it provides an optimal balance between horizontal and vertical components. However, depending on the context, such as initial height in this problem, slightly different angles may result in even better outcomes.

The angle of projection affects:

• The trajectory or path the projectile will follow
• The duration of the flight time
• The maximum height the projectile will reach

By adjusting the angle, different aspects of the trajectory, such as distance and height, can be optimized.

Initial Velocity

Initial velocity is the speed at which a projectile, like the hammer, starts its journey. It's made up of two components: one along the horizontal axis and the other along the vertical axis. These components are calculated using basic trigonometry, specifically involving sine and cosine functions.

For this hammer throw, the initial velocity is \(33\) feet per second. The horizontal and vertical components are determined using the formulas \(v_0 \cdot \cos(\theta)\) for the horizontal and \(v_0 \cdot \sin(\theta)\) for the vertical, where \(v_0\) is the initial speed and \(\theta\) the angle of projection.

Initial velocity is critical because:

• It dictates how far and fast the object will travel horizontally
• Influences the maximum height achieved
• Affects the time the projectile remains in the air

The balance between these components ultimately defines the object's flight path and influences how far it lands from the starting point.

Quadratic Formula

The quadratic formula is a mathematical tool used to find solutions to quadratic equations of the form \(ax^2 + bx + c = 0\). In projectile motion, it helps determine the time when the projectile, such as the hammer, will hit the ground.

In this exercise, the equation \(6 + (33 \sin(42^{\circ})) t - 16t^2 = 0\) is established by setting the vertical position function to zero, which represents the moment the hammer returns to ground level. Applying the quadratic formula \[t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] allows us to solve for the time \(t\) when this happens, where \(a=8\), \(b=33 \sin(42^{\circ})\), and \(c=-6\).

The quadratic formula is powerful because it:

• Provides precise values for the variables of interest
• Is applicable to various problems involving parabolic equations
• Enables the computation of key parameters without requiring graphical interpretation

Understanding its application in physics problems helps one to analytically determine the behavior of moving objects in relation to time and position.