Problem 61 - A Graphical Approach to Precalculus with Limits

Question

A projectile has been launched from the ground with an initial velocity of 88 feet per second. You are given parametric equations that model the path of the projectile. (a) Graph the parametric equations. (b) Approximate \(\theta\), the angle the projectile makes with the horizontal at launch, to the nearest tenth of a degree. (c) On the basis of your answer to part (b), write parametric equations for the projectile, using the cosine and sine functions. $$x=82.69265063 t, y=-16 t^{2}+30.09777261 t$$

Step-by-Step Solution

Answer

Launch angle \( \theta \approx 17.4^\circ \). New equations: \( x = 82.69 t, y = -16 t^2 + 26.46 t \).

Step 1: Identify given parametric equations

The parametric equations given are:\[ x = 82.69265063 t \]\[ y = -16 t^2 + 30.09777261 t \]

Step 2: Graph the parametric equations

Using a graphing calculator or software like Desmos, plot the parametric equations:- Set the range for \(t\) from 0 to the point where \(y = 0\) (the projectile lands).- You'll observe a parabolic trajectory, opening downwards, representing the path of the projectile.

Step 3: Determine horizontal component of velocity

The horizontal component of velocity is given by \( \frac{dx}{dt} = 82.69265063 \). This velocity is the product of the initial velocity and the cosine of the launch angle: \[ v_x = 88 \cos(\theta) = 82.69265063 \]

Step 4: Calculate the launch angle θ

By solving for \(\cos(\theta)\), we find:\[ \cos(\theta) = \frac{82.69265063}{88} \]Calculate \(\cos(\theta)\) and then use the inverse cosine function to find the angle:\[ \theta = \cos^{-1}\left(\frac{82.69265063}{88}\right) \approx 17.4^\circ \]

Step 5: Write new parametric equations using cosine and sine

With \( \theta \approx 17.4^\circ \), use the trigonometric identities to write new parametric equations:- \( x = 88 \cos(17.4^\circ) t \)- \( y = -16 t^2 + 88 \sin(17.4^\circ) t \)Using \(\cos(17.4^\circ) \approx 0.95630475596 \) and \(\sin(17.4^\circ) \approx 0.3007057995 \), you'll get:- \( x = 82.69 t \)- \( y = -16 t^2 + 26.46 t \)

Key Concepts

Projectile Motion

Projectile motion describes the curved path that an object follows when it is thrown or propelled near the earth's surface. In the given exercise, a projectile is launched with an initial velocity from the ground. As it moves through the air, the only significant forces acting upon it are gravity and its initial velocity.

The horizontal distance the projectile travels and its vertical displacement can both be described with parametric equations. These equations help us to understand how the projectile's position changes over time.

• The **horizontal motion** is governed by uniform velocity: there is no horizontal acceleration as there is no air resistance considered in this scenario.
• The **vertical motion** is influenced by gravity: this is an accelerated motion with constant acceleration due to gravity.

Parametric equations effectively separate these two components, where one variable (usually time) governs the change in both horizontal and vertical positions. The path of the projectile is a parabola, which reflects how gravity continually affects its vertical motion.

Trigonometry

Trigonometry comes into play when determining the direction and components of a projectile's initial velocity. Each projectile motion starts with a certain angle, called the launch angle, which affects the projectile's path significantly.

For the given exercise, the problem requires us to approximate the launch angle \( \theta \). Trigonometric functions, such as cosine and sine, are crucial for breaking down the projectile's initial velocity into horizontal and vertical components:

• The **cosine** of \( \theta \) is used to calculate the horizontal component, represented by \( v_x = v_0 \cos(\theta) \).
• The **sine** of \( \theta \) determines the vertical component, given by \( v_y = v_0 \sin(\theta) \).

The exercise uses the relationship between these components and the initial overall velocity to solve for the launch angle \( \theta \), using the inverse cosine function. The angle is found based on the ratio of horizontal velocity to the total initial velocity.

Velocity Calculations

Velocity calculations are essential in understanding how a projectile behaves. In this exercise, the velocity is decomposed into constant horizontal velocity and time-variable vertical velocity.

The horizontal velocity, \( v_x \), is constant throughout the motion because we assume no air resistance, providing a straightforward calculation: \( v_x = 82.69265063 \text{ feet per second} \). This is equivalent to calculating it using \( 88 \cos(\theta) \), where 88 is the initial velocity.

• To determine \( \theta \), the horizontal component formula \( v_x = 88 \cos(\theta) \) was rearranged to solve for \( \cos(\theta) \).
• Through this calculation, \( \theta \) is found to be approximately \( 17.4^\circ \).

The vertical component, involving gravity, causes a change in the vertical velocity over time. At any given point of time, the vertical part of the velocity changes due to the gravitational pull, which is reflected in the quadratic term \(-16t^2\) in the parametric equation for \( y \). This change explains the projectile's rise to a peak before descending back to the ground.