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Chapter 4: Problem 29

For whet launch angle will the height and garage of a projectile be equal?

Short Answer

Expert verified
The launch angle is \( \theta = \arctan(4) \).

Step by step solution

01

Understand the Problem

We need to find the launch angle of a projectile where the maximum height reached is equal to the range, or horizontal distance traveled, of the projectile.
02

Recall the Equations

The maximum height (\[ H \]) of a projectile launched with speed \( v \) at angle \( \theta \) is given by: \[ H = \frac{v^2 \sin^2\theta}{2g} \] where \( g \) is the acceleration due to gravity.The range (\[ R \]) is given by:\[ R = \frac{v^2 \sin(2\theta)}{g} \]
03

Set Height Equal to Range

Set the equation for maximum height equal to the equation for range: \[ \frac{v^2 \sin^2\theta}{2g} = \frac{v^2 \sin(2\theta)}{g} \]
04

Simplify the Equation

Cancel \( v^2/g \) from both sides of the equation:\[ \frac{\sin^2\theta}{2} = \sin(2\theta) \]
05

Use Trigonometric Identities

Use the identity \( \sin(2\theta) = 2\sin\theta\cos\theta \) and substitute it into the equation:\[ \frac{\sin^2\theta}{2} = 2\sin\theta\cos\theta \]
06

Solve for \( \theta \)

Simplify and rearrange the equation:\[ \sin^2\theta = 4\sin\theta\cos\theta \] Divide both sides by \( \sin\theta \) (assuming \( \theta eq 0 \) and \( \theta eq 180^\circ \) to avoid division by zero):\[ \sin\theta = 4\cos\theta \] Using \( \tan\theta = \frac{\sin\theta}{\cos\theta} \), we can write:\[ \tan\theta = 4 \] Thus, \( \theta = \arctan(4) \).

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Key Concepts

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

Launch angle
The launch angle, often denoted as \( \theta \), is a crucial factor when analyzing projectile motion. It determines how far and how high the projectile will go when launched. When we talk about the launch angle in this context, we're looking for the specific angle at which a projectile reaches maximum height equal to its range, or the total horizontal distance it travels.
  • The launch angle impacts both vertical and horizontal components of velocity, which in turn affects the projectile's trajectory and outcomes like range and height.
  • In most projectile motion problems, the launch angle is the adjustable factor that can be modified to achieve the desired result, such as maximizing the range or reaching a particular height.
In the context of our exercise, we aim to find that special launch angle where the projectile's highest point is the same as its entire horizontal travel distance. Analyzing how changing the angle influences these characteristics is essential for understanding projectile motion.
Trigonometric identities
Trigonometric identities play a vital role in solving projectile motion problems, especially when calculating the range and height of a projectile. They help simplify and solve the equations involving sin and cosine functions typically present in these problems.
  • The most relevant identity in this context is \( \sin(2\theta) = 2\sin\theta\cos\theta \), which directly connects the double angle to the sine and cosine values of a single angle.
  • This identity allowed us to replace \( \sin(2\theta) \) during simplification, aiding in finding the solution more easily.
Understanding these trigonometric identities is crucial because they often simplify the calculations, revealing surprising insights into the behavior of projectiles under different initial conditions, as further illustrated by finding \( \theta = \arctan(4) \).
Range and height equations
In projectile motion, two primary equations describe how far a projectile will travel horizontally (range) and how high it will go (height). Understanding how to use and manipulate these equations is essential for solving related problems.
  • The **range equation** is \( R = \frac{v^2 \sin(2\theta)}{g} \), indicating the horizontal distance traveled. Here, \( v \) represents the launch speed and \( g \) is gravitational acceleration.
  • Meanwhile, the **maximum height equation** is \( H = \frac{v^2 \sin^2\theta}{2g} \), describing the peak vertical position achieved by the projectile.
In our exercise, these equations were set equal to discern the special angle \( \theta \) that makes the height equal to the range, illuminating a unique scenario in projectile dynamics. Mastering these equations allows for deeper insight and more precise predictions in real-world physics applications.

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Most popular questions from this chapter

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