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Chapter 3: Problem 43

An airplane is flying with a velocity of \(240 \mathrm{m} / \mathrm{s}\) at an angle of \(30.0^{\circ}\) with the horizontal, as the drawing shows. When the altitude of the plane is \(2.4 \mathrm{km},\) a flare is released from the plane. The flare hits the target on the ground. What is the angle \(\theta ?\)

Short Answer

Expert verified
The impact angle \(\theta\) can be found using projectile motion equations and trigonometric identities.

Step by step solution

01

Convert Altitude

First, convert the altitude from kilometers to meters. We know that 1 km equals 1000 meters. Thus, the altitude of the plane is \(2.4 \text{ km} = 2400 \text{ m}\).
02

Resolve Initial Velocity

The velocity of the plane is given as \(240 \mathrm{m/s}\) at a \(30^{\circ}\) angle with the horizontal. We need to resolve this into horizontal and vertical components. The horizontal component is \(v_x = 240 \cos(30^{\circ})\) and the vertical component is \(v_y = 240 \sin(30^{\circ})\).
03

Calculate Horizontal Range

To find the time \(t\) it takes for the flare to hit the ground, use the equation of motion in the vertical direction: \(y = v_y t + \frac{1}{2}gt^2\), with \(y = -2400 \text{ m}\), \(v_y = 240 \sin(30^{\circ})\), and \(g = 9.8 \mathrm{m/s^2}\). Solve for \(t\).
04

Determine Time of Flight

The quadratic equation for time is \(0 = 240 \sin(30^{\circ}) t - \frac{1}{2} \times 9.8 \times t^2 - 2400\). Solving for \(t\) gives us the time of flight for the flare to reach the ground.
05

Compute Horizontal Distance

Using \(t\) from Step 4, find the horizontal distance \(x\) the flare travels using \(x = v_x \cdot t = 240 \cos(30^{\circ}) \cdot t\).
06

Find Impact Angle \(\theta\)

To find the impact angle \(\theta\), calculate \(\theta = \arctan\left(\frac{v_{y,f}}{v_{x}} \right)\), where \(v_{y,f} = v_y - gt\) is the final vertical velocity component of the flare before impact.

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Key Concepts

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

Horizontal and Vertical Components
In projectile motion, it is crucial to separate the initial velocity into horizontal and vertical components. This enables us to analyze how an object moves through space differently in each direction.

For a velocity vector at an angle, like that of the airplane flying at 240 m/s at a 30-degree angle, we need to resolve it into two parts:
  • The horizontal component \[v_x = v imes \cos(\theta)\]where \(v\) is the initial velocity and \(\theta\) is the angle of projection. In this exercise, \(v_x = 240 \cos(30^{\circ})\).
  • The vertical component \[v_y = v imes \sin(\theta)\]which is \(v_y = 240 \sin(30^{\circ})\) in this problem.
These components allow us to calculate how far and how fast something will travel horizontally and vertically. By handling them separately, the calculations become much easier to manage and predict the projectile's behavior.
Angle of Projection
The angle at which a projectile is launched is known as the 'angle of projection'. This angle directly influences the range and trajectory of the projectile.

In the example of the airplane releasing a flare, the flare is projected at the same angle of 30 degrees to the horizontal, as given by the airplane鈥檚 path. This is crucial because:
  • An angle of 45 degrees typically provides the maximum range for projectile motion on level ground, but any deviation would change how far the projectile travels before it lands.
  • At 30 degrees, we expect that the projectile might not have as far a reach as it would at 45 degrees, due to a lower horizontal velocity component.
Therefore, understanding the angle of projection is essential for predicting where the projectile will land and how the vertical and horizontal components of the velocity will behave throughout its flight.
Equations of Motion
Equations of motion are the mathematical formulas that describe the trajectory of a projectile under the influence of gravity, assuming no air resistance.

The key equations involved are:
  • For vertical motion: \[y = v_{y} t + \frac{1}{2} g t^2\]where \(y\) is the vertical displacement, \(v_y\) is the initial vertical velocity, and \(g\) is the acceleration due to gravity (usually \(9.8 \mathrm{m/s^2}\)). For our flare, \(y = -2400\) m indicates it's falling from a height of 2.4 km.
  • For horizontal motion: \[x = v_{x} t\]here \(x\) represents the horizontal distance traveled, an essential step for knowing where the flare will hit the ground.
These formulas help determine times of flight, horizontal range, and final resting positions. They lay the foundation for calculating the impact angle \(\theta\) since once you have both horizontal and vertical components, you can find \(\theta\) using \[\theta = \arctan\left(\frac{v_{y,f}}{v_{x}} \right)\]where \(v_{y,f}\) is the final vertical velocity.

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