Problem 59 - A Graphical Approach to Precalculus with Limits

Question

If an object is projected on the moon, then the parametric equations of flight are $$ x=(v \cos \theta) t \quad \text { and } \quad y=(v \sin \theta) t-2.66 t^{2}+h $$ Estimate the distance that a golf ball hit at 88 feet per second \((60 \mathrm{mph})\) at an angle of \(45^{\circ}\) with the horizontal travels on the moon if the moon's surface is level.

Step-by-Step Solution

Answer

The golf ball travels approximately 1455 feet on the moon's level surface.

Step 1: Understanding the Problem

We need to calculate the distance a golf ball travels on the moon, given that its speed is 88 feet per second, it is projected at a 45-degree angle, and the initial height \( h \) is considered from the moon's surface. The distance is determined by the horizontal trajectory of the ball.

Step 2: Breaking Down the Given Equations

The parametric equations are \( x = (v \cos \theta) t \) and \( y = (v \sin \theta) t - 2.66 t^2 + h \). Here, \( v = 88 \) ft/s and \( \theta = 45^\circ \).

Step 3: Substitute Known Values

Using \( v = 88 \) ft/s and \( \theta = 45^\circ \), the horizontal equation becomes \( x = (88 \cos 45^\circ) t \), and the vertical equation becomes \( y = (88 \sin 45^\circ) t - 2.66 t^2 + h \).

Step 4: Calculate Trigonometric Components

\( \cos 45^\circ = \sin 45^\circ = \frac{\sqrt{2}}{2} \), thus \( 88 \cos 45^\circ = 44\sqrt{2} \) and similarly \( 88 \sin 45^\circ = 44\sqrt{2} \). Update the equations: \( x = 44\sqrt{2} t \) and \( y = 44\sqrt{2} t - 2.66 t^2 + h \).

Step 5: Determine Time to Reach Ground

Set \( y = 0 \) and solve \( 0 = 44\sqrt{2} t - 2.66 t^2 + h \) (assuming \( h = 0 \)).Factor the equation: \(-2.66 t^2 + 44\sqrt{2} t = 0\).Use quadratic formula or factor to find \( t = 0 \) (at launch) or \( t = \) non-zero time when it hits the ground.

Step 6: Solve for Non-zero Time

Solve \( -2.66 t^2 + 44\sqrt{2} t = 0 \).First, factor out \( t \): \( t(-2.66 t + 44\sqrt{2}) = 0 \).The non-zero solution is \( t = \frac{44\sqrt{2}}{2.66} \).

Step 7: Calculate the Horizontal Distance

Substitute the non-zero value of \( t \) back into the horizontal equation \( x = 44\sqrt{2} t \) to find the distance traveled, \( x = 44\sqrt{2} \times \frac{44\sqrt{2}}{2.66} \).

Step 8: Final Calculation

Perform the arithmetic: convert \( 44\sqrt{2} \) into a numerical value, multiply by itself, and divide by \( 2.66 \). The approximate distance is calculated.

Key Concepts

Trigonometric Functions

Trigonometry plays a crucial role in solving problems involving angles and distances, such as in the given parametric equations. The trigonometric functions, specifically sine and cosine, help us determine the horizontal and vertical components of a projectile's motion. In our problem, the angle of projection is given as \( 45^{\circ} \), which is a common angle in trigonometry, often simplifying calculations due to its special properties:

• \( \cos 45^{\circ} = \sin 45^{\circ} = \frac{\sqrt{2}}{2} \)

These values help break down any vector into its effective horizontal and vertical components. For instance, if a projectile is launched at a speed \( v \) and an angle \( \theta \), the horizontal component is determined by \( v \cos \theta \) and the vertical component by \( v \sin \theta \).
In our exercise, this decomposition aids in calculating both the maximum height reached and the time of flight by manipulating these basic trigonometric functions. Understanding these concepts can simplify complex trajectories into manageable calculations. Remember that 45-degree angles are very helpful because they create equal horizontal and vertical components of the motion.

Projectile Motion

Projectile motion is a fascinating concept in physics that involves the motion of an object projected into the air, subject only to the acceleration due to gravity. The object's path is typically a parabola shaped by its initial velocity and the angle of projection. In the exercise, we are dealing with the physics of projectile motion without air resistance and on the moon, where gravity is less intense than Earth.
When solving projectile motion problems, the path can be characterized by two parametric equations: one for the horizontal motion \( x \) and one for the vertical motion \( y \). These parametric equations take into account the object's initial velocity, angle of projection, and gravity:

• The horizontal motion equation is \( x = (v \cos \theta) t \), reflecting constant velocity due to the absence of horizontal forces.
• The vertical motion equation is \( y = (v \sin \theta) t - at^2 + h \), where \( a \) represents the gravitational acceleration on the moon.

These equations illustrate important aspects of motion, where gravity only affects vertical speed. Understanding how gravity interacts with a projectile's vertical velocity is crucial to predicting where and when it will land.
In our problem, solving for when the projectile hits the ground involves setting \( y = 0 \) and solving the resulting quadratic equation for time \( t \), which we can then use to calculate the distance traveled horizontally.

Quadratic Equations

Quadratic equations are equations of the form \( ax^2 + bx + c = 0 \) and are pivotal in finding solutions to problems involving parabolic trajectories in projectile motion. In our case, the vertical motion of the projectile is described by a quadratic equation due to the gravitational term \( -2.66 t^2 \).
To find out when the golf ball hit the ground on the moon, we need to solve for \( t \) in the equation \( y = (v \sin \theta) t - 2.66 t^2 + h = 0 \). Often, this involves:

• Rearranging the equation into standard quadratic form \( at^2 + bt + c = 0 \)
• Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) if the equation cannot be factored easily

In this practical application, the non-zero solution of \( t \) corresponds to when the projectile touches the ground. Solving this accurately is key to understanding the duration of the projectile's flight, which in turn allows us to calculate its range.
Working with quadratic equations in physics helps us model and predict motion, providing critical insights into how objects move under various forces. The solution will not only tell us when the object reaches the ground but also how far it has traveled horizontally. Understanding these equations and their solutions is fundamental to mastering projectile motion problems.