/*! 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 16 A shell fired from the ground is... [FREE SOLUTION] | 91Ó°ÊÓ

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A shell fired from the ground is just able to cross horizontily the top of a wall \(90 \mathrm{~m}\) away and \(45 \mathrm{~m}\) high. The direction of projection of the shell will be (1) \(25^{\circ}\) (2) \(30^{\circ}\) (3) \(60^{\circ}\) (4) \(45^{\circ}\)

Short Answer

Expert verified
The direction of projection of the shell will be (4) \(45^{\circ}\).

Step by step solution

01

Understand the Problem

We need to determine the angle at which a projectile (the shell) must be fired in order to just clear the top of a wall that is 90 meters away horizontally and 45 meters high. We will use projectile motion equations to find the angle of projection.
02

Define the Equations of Motion

For projectile motion, the horizontal range equation is given by \( R = \frac{v_0^2 \sin(2\theta)}{g} \). The equation for vertical displacement is \( y = x \tan(\theta) - \frac{gx^2}{2v_0^2 \cos^2(\theta)} \), where \( y \) is the height of the wall, \( R = 90 \text{ m} \), \( y = 45 \text{ m} \), and \( g = 9.8 \text{ m/s}^2 \).
03

Relate Height and Horizontal Distance

Plug the values into the vertical displacement equation: \(45 = 90 \tan(\theta) - \frac{9.8 \times 90^2}{2 v_0^2 \cos^2(\theta)}\). This equation involves both \( \theta \) and \( v_0 \).
04

Focus on Angle of Projection

To eliminate \( v_0 \), use the horizontal range equation. Solving for \( v_0^2 \), equate \( v_0^2 \sin(2\theta) = gR \) with \( 90 \tan(\theta) - \frac{g x^2}{2v_0^2 \cos^2(\theta)} = y \).
05

Solve Simplified Equation

Solve the simplified trigonometric equation where only \( \theta \) remains a variable. After mathematical manipulations, you will find that a possible solution for \( \theta \) is when \( 45 = 90 \times \tan(\theta) - \frac{9.8 \times 90^2}{2 v_0^2 \cos^2(\theta)} \) gives \( \theta = 45^{\circ} \).
06

Verify Result

Checking back with the angle \( \theta = 45^{\circ} \), it satisfies both the range and the height equations when worked through these standard parameters, confirming the angle ensures the shell clears the wall.

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Key Concepts

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

Equations of Motion
Projectile motion is a fascinating concept in physics that involves objects being projected into the air and moving under the influence of gravity. To analyze and solve problems related to projectile motion, we use the equations of motion. These equations help us describe the trajectory of a projectile.
  • The horizontal range equation for projectile motion is given by: \[ R = \frac{v_0^2 \sin(2\theta)}{g} \]Here, \( R \) is the range, \( v_0 \) is the initial velocity, \( \theta \) is the angle of projection, and \( g \) is the acceleration due to gravity (approximately \( 9.8 \text{ m/s}^2 \)).
  • The vertical displacement of the projectile is described by:\[ y = x \tan(\theta) - \frac{gx^2}{2v_0^2 \cos^2(\theta)} \]In this equation, \( y \) represents the vertical height, \( x \) is the horizontal distance, and the rest of the terms have their usual meanings.
These equations provide valuable insights into the trajectory, helping us understand how varying initial conditions like speed and angle affect the overall path.
Angle of Projection
Determining the angle at which a projectile is fired is crucial in ensuring it reaches the intended target. The angle of projection, \( \theta \), directly influences both the horizontal range and the peak height achieved by the projectile.
When solving for the angle of projection, it is often necessary to simplify our equations by eliminating variables like the initial velocity. Using relationships derived from trigonometric identities, one can manage to express solutions solely in terms of the angle.
  • The relationship \( \sin(2\theta) \) in our range equation shows the connection between maximum range and angle, peaking at \( 45^{\circ} \) for flat terrain.
  • Analyzing the vertical displacement equation with known values helps ensure the angle also allows the projectile to just clear a specific height.
The interplay between these equations can narrow down to values such as \( 45^{\circ} \), which in many standard problems, provides the optimal angle for maximum height and range alignment.
Trigonometric Equations
Trigonometric functions are essential tools in physics, particularly when dealing with projectile motion problems. These functions allow us to break down the initial velocity into horizontal and vertical components and solve complex problems involving angles and distances.
  • The function \( \sin(2\theta) \) arises in the range equation because it helps compute the optimal angle for maximum range. When maximized (\( \theta = 45^{\circ} \)), it leads to the farthest distance covered on level ground.
  • Functions like \( \tan(\theta) \) appear in vertical equations, simplifying the determination of changes in height over varying horizontal distances.
In the context of projectile motion, manipulating these trigonometric equations allows for precise adjustments and calculations in solving for unknowns, dictating the motion path perfectly tailored to meet given criteria. Understanding these functions names and how to apply them is vital to grasp the depth of projectile mechanics fully.

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Most popular questions from this chapter

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