Problem 49 - University Physics with Modern Physics

Question

An airplane is flying with a velocity of 90.0 m/s at an angle of 23.0\(^{\circ}\) above the horizontal. When the plane is 114 m directly above a dog that is standing on level ground, a suitcase drops out of the luggage compartment. How far from the dog will the suitcase land? Ignore air resistance.

Step-by-Step Solution

Answer

The suitcase lands about 758.67 meters from the dog.

Step 1: Break Down the Problem

The suitcase is dropped from an airplane flying at an angle, meaning it has both horizontal and vertical components of initial velocity. Our goal is to find how far the suitcase travels horizontally (distance from the dog) when it hits the ground.

Step 2: Determine the Initial Velocity Components

The initial velocity of the suitcase, which is the same as the airplane's velocity, is 90.0 m/s. To find the horizontal and vertical components of this velocity, we use trigonometric functions: \( v_{x} = v \cos(\theta) \) and \( v_{y} = v \sin(\theta) \). We have \( v_{x} = 90.0 \cos(23.0^{\circ}) \approx 82.65 \text{ m/s} \) and \( v_{y} = 90.0 \sin(23.0^{\circ}) \approx 35.12 \text{ m/s} \).

Step 3: Calculate the Time of Flight

Using the vertical motion equations, where the initial vertical position \( y_0 = 114 \text{ m} \) and the vertical velocity \( v_{y} \approx 35.12 \text{ m/s} \), determine the time of flight \( t \). Use \( y = y_0 + v_{y}t - \frac{1}{2}gt^2 \), setting \( y = 0 \) when the suitcase hits the ground. Solving \( 0 = 114 + 35.12t - \frac{1}{2}(9.8)t^2 \) leads to a quadratic equation \( 4.9t^2 - 35.12t - 114 = 0 \).

Step 4: Solve for Time Using the Quadratic Formula

The quadratic formula is \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 4.9 \), \( b = -35.12 \), and \( c = -114 \). Solving gives us \( t = \frac{35.12 \pm \sqrt{35.12^2 + 4 \times 4.9 \times 114}}{9.8} \approx 9.18 \text{ seconds} \).

Step 5: Calculate the Horizontal Distance

With the time of flight \( t = 9.18 \text{ seconds} \) and horizontal velocity \( v_{x} \approx 82.65 \text{ m/s} \), determine the horizontal distance \( d_x \) the suitcase travels: \( d_x = v_{x} \times t = 82.65 \times 9.18 \approx 758.67 \text{ meters} \).

Step 6: Conclude the Solution

The suitcase lands approximately 758.67 meters from the dog, following its horizontal trajectory, assuming no air resistance.

Key Concepts

Initial Velocity Components

In projectile motion problems, like the one involving the suitcase dropped from an airplane, understanding velocity components is fundamental. The initial velocity of the object, here the suitcase, must be split into horizontal and vertical components to analyze the motion accurately.

Given a velocity and an angle above the horizontal, trigonometric functions come into play to determine these components:

• The horizontal component, often labeled as \( v_{x} \), is found using \( v_{x} = v \cos(\theta) \). In the exercise, the suitcase's horizontal velocity is calculated as approximately 82.65 m/s.
• The vertical component, \( v_{y} \), is calculated using \( v_{y} = v \sin(\theta) \), resulting in around 35.12 m/s for the vertical speed of the suitcase.

These components allow us to separate the motion into manageable parts, enabling the calculation of how far the suitcase travels horizontally and how long it takes to reach the ground.

Time of Flight

The time of flight is a crucial element in projectile motion, representing the total time an object spends in the air. For vertical motion, this is determined by setting the vertical displacement equal to zero, indicating that the suitcase has hit the ground.

By using the equation for vertical motion:

• \( y = y_0 + v_{y}t - \frac{1}{2}gt^2 \), where \( y_0 \) is the initial vertical position, \( v_{y} \) is the vertical component of the initial velocity, and \( g \) is the acceleration due to gravity.

Setting \( y = 0 \) (the suitcase hits the ground), the exercise forms a quadratic equation which is then solved to find the time of flight, approximately 9.18 seconds in this case.

Horizontal Distance

Once the time of flight is found, calculating how far the suitcase travels horizontally involves the initial horizontal velocity component. This part of the motion is independent of the vertical motion since no horizontal forces act on the suitcase (neglecting air resistance).

Using the formula:

• \( d_x = v_{x} \times t \), where \( d_x \) is the horizontal distance, \( v_{x} \) the horizontal velocity, and \( t \) the time of flight.

This formula provides the horizontal distance the suitcase travels. Given \( v_{x} = 82.65 \) m/s and \( t = 9.18 \) seconds, the suitcase lands approximately 758.67 meters from the starting point, far from the dog standing below the drop point.

Quadratic Equation

The quadratic equation becomes essential in solving for the time of flight when analysing projectile motion. It stems from the equation for vertical displacement and is indispensable when an object is dropped from a height with initial vertical velocity.

The form of the quadratic equation used here is:

• \( 0 = 114 + 35.12t - \frac{1}{2}(9.8)t^2 \)

This is simplified to the standard quadratic format \( 4.9t^2 - 35.12t - 114 = 0 \). By applying the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), the roots of the equation represent the time it takes for the suitcase to reach the ground. The solution yields a time of approximately 9.18 seconds.

Trigonometric Functions

Trigonometric functions, particularly sine and cosine, are incredibly useful in projectile motion problems. They allow us to break down a velocity at an angle into horizontal and vertical components.

In this exercise:

• Using \( \cos(\theta) \) helps find the horizontal component \( v_{x} \), as in \( v_{x} = v \cos(23.0^{\circ}) \), resulting in approximately 82.65 m/s.
• Using \( \sin(\theta) \) helps find the vertical component \( v_{y} \), as in \( v_{y} = v \sin(23.0^{\circ}) \), resulting in roughly 35.12 m/s.

These functions help dissect the angled motion of the suitcase into simpler linear motions, allowing further calculations to determine the flight path and impact point efficiently.