/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 66 Projectile explorations A projec... [FREE SOLUTION] | 91Ó°ÊÓ

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Projectile explorations A projectile launched from the ground with an initial speed of \(20 \mathrm{m} / \mathrm{s}\) and a launch angle \(\theta\) follows a trajectory approximated by $$x=(20 \cos \theta) t, \quad y=-4.9 t^{2}+(20 \sin \theta) t$$ where \(x\) and \(y\) are the horizontal and vertical positions of the projectile relative to the launch point (0,0) a. Graph the trajectory for various values of \(\theta\) in the range \(0<\theta<\pi / 2\) b. Based on your observations, what value of \(\theta\) gives the greatest range (the horizontal distance between the launch and landing points \() ?\)

Short Answer

Expert verified
Answer: The optimal launch angle is \(\dfrac{\pi}{4}\) radians or \(45\) degrees.

Step by step solution

01

Determine the parametric equations of the path

The projectile trajectory is given by two parametric equations: \(x(t) = (20 \cos \theta) t\) for horizontal position and \(y(t) = -4.9 t^{2} + (20 \sin \theta) t\) for vertical position.
02

Vary \(\theta\)

To graph the trajectory for different values of \(\theta\), we can use a graphing software to plot the parametric equations. Let \(\theta\) vary in the range of \(0\) to \(\dfrac{\pi}{2}\) with increments of, for example, \(\dfrac{\pi}{8}\).
03

Graph

Insert the parametric equations into the graphing software with the defined range of \(\theta\) and generate a graph, which will show the trajectory of the projectile for various launch angles. b. Find the value of \(\theta\) that maximizes the range
04

Understand Range

The range of the projectile is the horizontal distance from the launch point to the landing point. Mathematically, since range is the horizontal distance, we're looking for the maximum distance of \(x\) when \(y = 0\).
05

Determine the time when the projectile lands

Since the landing point has a vertical position of \(0\), set the \(y(t)\) equation equal to zero: $$0 = -4.9 t^{2} + (20 \sin \theta) t$$ We can solve for the time \(t\) at which the projectile will land. Note that $$t = 0$$ is when the projectile is launched.
06

Solve for time and find the range equation

Factor out \(t\) and solve the equation for the time of landing, \(t_{land}\) $$0 = t(-4.9 t + (20 \sin \theta))$$ $$t_{land} = \frac{20 \sin \theta}{4.9}$$ Now we can find the range by plugging \(t_{land}\) into the \(x(t)\) function, which gives: $$x(t_{land}) = (20 \cos \theta) \frac{20 \sin \theta}{4.9}$$ $$\implies R(\theta) = 4.08\sin \theta \cos \theta$$
07

Maximize the range equation

To find the angle that maximizes the range, we can take the derivative of the range equation with respect to \(\theta\) and set it equal to zero: $$\frac{dR(\theta)}{d\theta} = 0$$ Using the double angle trigonometric identity, express \(R(\theta)\) in terms of \(\sin(2\theta)\): $$R(\theta) = 2.04 \sin(2\theta)$$ Differentiate the range function with respect to \(\theta\): $$\frac{dR(\theta)}{d\theta} = 2.04 (2 \cos(2\theta))$$ Setting the derivative to zero, we have: $$2.04(2 \cos(2\theta)) = 0$$ So, $$\cos(2\theta) = 0$$
08

Find the optimal launch angle

The equation \(\cos(2\theta) = 0\) gives us: $$2\theta = \frac{\pi}{2}$$ Simplifying, we get: $$\theta = \frac{\pi}{4}$$ Thus, the optimal launch angle \(\theta\) that gives the greatest range is \(\dfrac{\pi}{4}\) radians or \(45\) degrees.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Parametric Equations
In projectile motion, parametric equations play a crucial role in describing the trajectory of an object. They break down the complex path of a projectile into simpler, more manageable forms. Instead of one single equation, we use two separate equations to describe the horizontal and vertical positions of a projectile over time.

For our projectile, the path is described using the parametric equations:
  • Horizontal position: \(x(t) = (20 \cos \theta) t\)
  • Vertical position: \(y(t) = -4.9 t^2 + (20 \sin \theta) t \)
In these equations, \(t\) represents time, and \(\theta\) is the launch angle. The parameter \(t\) allows us to describe the time evolution of the positions, making it possible to model real-world physics problems like the motion of a projectile with ease. This time-driven approach offers a clear method for visualizing how the projectile moves through space.

Overall, parametric equations allow us to separate the effects of time on both horizontal and vertical motion, which is particularly helpful when analyzing trajectories.
Launch Angle
The launch angle \(\theta\) is a critical parameter in determining a projectile's path. It influences both the range and the peak height of the projectile. In essence, the launch angle dictates how steeply the projectile is fired.

Consider the projectile's parametric equations:
  • \(x(t) = (20 \cos \theta) t\)
  • \(y(t) = -4.9 t^2 + (20 \sin \theta) t \)
The angle affects the initial velocity components, specifically:
  • The horizontal component \(20 \cos \theta\)
  • The vertical component \(20 \sin \theta\)
A launch angle \(\theta\) of 45 degrees, or \(\frac{\pi}{4}\) radians, is found to maximize the range of the projectile. At this angle, the energy is optimally split between horizontal distance and peak height.

The importance of the launch angle is visible when graphing different trajectories. As \(\theta\) varies, the trajectory changes shape. Lower angles yield longer, flatter paths, while steeper angles launch the projectile higher but closer. Balancing height and distance is key to maximizing range.
Trigonometric Functions
Trigonometric functions underpin the calculations of projectile motion. They allow us to resolve the initial velocity into its horizontal and vertical components, ensuring accurate modeling of the trajectory.

For instance, in our example:
  • The horizontal component: \( v_x = 20 \cos \theta \)
  • The vertical component: \( v_y = 20 \sin \theta \)
The functions \(\cos \theta\) and \(\sin \theta\) help distribute the speed \(20 \mathrm{m/s}\) in different directions based on the angle \(\theta\). When solving for optimal angles, we employ identities like the double angle formula: \(\sin(2\theta) = 2 \sin \theta \cos \theta\).

This understanding is essential when optimizing aspects like range. In the provided solution, utilizing \( R(\theta) = 2.04 \sin(2\theta) \) guides us to determine that the range is maximum when \(\theta = \frac{\pi}{4}\). Trigonometry thus not only aids in computing trajectories but also in optimizing paths for desired outcomes.

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