Problem 30 - College Physics

Question

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

Step-by-Step Solution

Answer

The suitcase lands approximately 712.34 meters away from the dog.

Step 1: Break Down Components of Initial Velocity

The plane is flying at a velocity of 90.0 m/s at an angle of 23.0° above the horizontal. We need to find the horizontal and vertical components of this velocity.The horizontal component \( v_{x} \) is given by \( v_{x} = v \cos \theta \), and the vertical component \( v_{y} \) is given by \( v_{y} = v \sin \theta \).Substitute the given values:\[ v_{x} = 90.0 \cos 23.0^{\circ} \approx 82.83 \text{ m/s} \]\[ v_{y} = 90.0 \sin 23.0^{\circ} \approx 35.15 \text{ m/s} \]

Step 2: Calculate the Time of Flight

Since the suitcase is dropped from a height of 114 m, we can use the vertical motion formula for objects under gravity to find the time it takes for the suitcase to hit the ground. The formula is:\[ h = v_{y} t + \frac{1}{2} g t^{2} \]where \( h = 114 \text{ m} \), \( v_{y} = 35.15 \text{ m/s} \), and \( g = -9.81 \text{ m/s}^2 \). Rearrange the equation and solve for time \( t \):\[ 0 = 114 + 35.15t - 4.905t^2 \]This is a quadratic equation in the form \( at^2 + bt + c = 0 \). Use the quadratic formula:\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Where \( a = -4.905 \), \( b = 35.15 \), and \( c = 114 \). Solve to find time \( t \approx 8.60 \text{ seconds} \).

Step 3: Calculate the Horizontal Distance Traveled

With the time of flight calculated in Step 2, we now determine how far the suitcase travels horizontally while it is falling.The horizontal distance \( d_x \) is given by:\[ d_x = v_x \cdot t \]Substitute the known values:\[ d_x = 82.83 \text{ m/s} \cdot 8.60 \text{ s} \approx 712.34 \text{ m} \]

Step 4: Conclude the Distance from the Dog to Suitcase Landing Point

Since the suitcase falls horizontally starting 114 meters above the dog, and no horizontal forces act on it (ignoring air resistance), the calculated horizontal displacement \( d_x \) is the distance from the dog to where the suitcase lands.Thus, the suitcase lands approximately 712.34 meters away from the dog.

Key Concepts

Vertical Motion

Vertical motion describes how the suitcase falls after it leaves the airplane. Since gravity is the only force acting on it vertically, the suitcase undergoes constant acceleration downwards at approximately 9.81 m/s².
This is known as free fall. To calculate how long it takes the suitcase to reach the ground, we use formulas from kinematics.
This involves the initial vertical velocity and the height from which it was dropped.

• Initial vertical velocity is calculated from the velocity of the plane and the angle at which it flies.
• The height is the initial vertical distance between the plane and the ground.

For this suitcase, the equation involves replacing into the quadratic formula to solve for time. This calculation results in a time of about 8.60 seconds.

Horizontal Motion

While the suitcase is falling vertically, its horizontal motion is straightforward. Since there are no horizontal forces acting on the suitcase (ignoring air resistance), it maintains a constant velocity. In our exercise, this velocity is derived from the horizontal component of the airplane's initial velocity.
The horizontal distance traveled only depends on this constant horizontal velocity and the duration it takes for the suitcase to fall, also known as the time of flight.

• The longer the time of flight, the farther the suitcase will travel horizontally.
• With a calculated horizontal speed of about 82.83 m/s, the suitcase covers more distance over more time.

Hence, the overall motion is the combination of falling vertically under gravity and moving horizontally at a steady pace.

Initial Velocity Components

To accurately determine the movement of the suitcase upon release, it is crucial to break down its initial velocity into vertical and horizontal components. This separation of components is foundational in understanding projectile motion.
Let's explore how to calculate these components:

• **Horizontal Component**: This is calculated using the formula \( v_x = v \cos \theta \), where \( v \) is the initial velocity of the plane, and \( \theta \) is the angle of projection. For the given problem, this component is approximately 82.83 m/s.
• **Vertical Component**: Similarly, this is determined using \( v_y = v \sin \theta \). For our suitcase, it turns out to be around 35.15 m/s.

By breaking the velocity into these two components, we can separately analyze how the suitcase moves vertically and horizontally.

Time of Flight

The time of flight is the time the suitcase spends in the air from the moment it drops until it hits the ground. In the case of the suitcase, it's affected primarily by the vertical motion under gravity. Calculating this requires solving a quadratic equation obtained from the vertical motion formula:
\[ h = v_y t + \frac{1}{2} g t^2 \] Here, \( h \) is the initial height (114 meters), \( v_y \) is the vertical component of initial velocity, and \( g \) is the acceleration due to gravity.
The quadratic formula is then used to solve for \( t \), resulting in approximately 8.60 seconds.

• During this time, the suitcase falls and covers a distance horizontally at the initial horizontal velocity.
• Hence, understanding the time of flight helps in predicting both vertical landing point and horizontal range.

Consequently, the suitcase lands far from its initial drop point due to its steady horizontal speed over 8.60 seconds.