Problem 63 - Physics for Scientists and Engineers with Modern Physics

Question

(II) A person in the passenger basket of a hot-air balloon throws a ball horizontally outward from the basket with spced 10.0 \(\mathrm{m} / \mathrm{s}\) (Fig. 52\()\) . What initial velocity (magnitude and direction) does the ball have relative to a person standing on the ground (a) if the hot-air balloon is rising at 5.0 \(\mathrm{m} / \mathrm{s}\) relative to the ground during this throw, (b) if the hot-air balloon is descending at 5.0 \(\mathrm{m} / \mathrm{s}\) relative to the ground.

Step-by-Step Solution

Answer

(a) 11.2 m/s at 26.6° above horizontal; (b) 11.2 m/s at 26.6° below horizontal.

Step 1: Understand the Problem

To find the initial velocity of the ball relative to a person standing on the ground, we must account for both the horizontal speed of the throw and the vertical speed of the balloon. We'll do this for two scenarios: when the balloon is rising and when it's descending.

Step 2: Set Up the Known Quantities

The ball is thrown with a horizontal speed of 10.0 m/s. When the balloon is rising, its vertical speed is +5.0 m/s, and when descending, its vertical speed is -5.0 m/s.

Step 3: Calculate the Relative Velocity in Scenario A (Balloon Rising)

For scenario (a), the horizontal velocity of the ball remains 10.0 m/s. The vertical velocity relative to the ground is 5.0 m/s as well. Using the Pythagorean theorem \(v = \sqrt{v_x^2 + v_y^2}\), the magnitude of the velocity is \(\sqrt{10.0^2 + 5.0^2} = \sqrt{100 + 25} = \sqrt{125} \approx 11.2 \, \text{m/s}\).

Step 4: Determine the Direction (Angle) for Scenario A

The angle \(\theta\) with the horizontal is given by \(\tan(\theta) = \frac{v_y}{v_x}\). Thus, \(\theta = \tan^{-1}\left(\frac{5.0}{10.0}\right) = \tan^{-1}(0.5) \approx 26.6^\circ\).

Step 5: Calculate the Relative Velocity in Scenario B (Balloon Descending)

Now, consider when the balloon is descending. The horizontal velocity remains at 10.0 m/s, but the vertical velocity is -5.0 m/s. The magnitude of the velocity is calculated as: \(\sqrt{10.0^2 + (-5.0)^2} = \sqrt{100 + 25} = \sqrt{125} \approx 11.2 \, \text{m/s}\).

Step 6: Determine the Direction (Angle) for Scenario B

For scenario (b), the angle \(\theta\) is given by \(\theta = \tan^{-1}\left(\frac{-5.0}{10.0}\right) = \tan^{-1}(-0.5) \approx -26.6^\circ\), which indicates the direction below the horizontal.

Key Concepts

Relative Velocity

Relative velocity is an essential concept in understanding projectile motion. It deals with how the velocity of one object appears from the reference frame of another object. In this problem, you're dealing with two perspectives: the perspective of the person in the hot-air balloon and the person standing on the ground.

• When an object is thrown from a moving vehicle or platform, like a hot-air balloon, its velocity relative to another observer must account for both the speed of the vehicle and how fast the object is thrown.
• For understanding relative velocity, you can think of it as the vector addition of the object's velocity and the vehicle's velocity.

In a mathematical sense, when the hot-air balloon is rising, the relative velocity of the ball to someone on the ground is the vector sum of the horizontal throwing speed and the upward speed of the balloon.
When the balloon is descending, the vertical component changes direction. Therefore, the relative velocity still involves the horizontal component, but the vertical component is now downwards. Relative velocity can help you understand why the angles in the two scenarios are positioned differently around the horizontal line.

Hot-Air Balloon Dynamics

Hot-air balloon dynamics are crucial for interpreting how movement impacts projectile motion. In physics, analyzing balloon movement involves understanding how upward and downward velocities affect objects released from them.

• In this exercise, the balloon's motion adds a vertical component to any object released from it. Whether the balloon moves up or down influences the initial velocity of the object relative to an observer on the ground.
• Think of the hot-air balloon's movement as a baseline velocity that alters how other velocities are perceived.

The balloon's dynamics follow basic principles of motion:

• If the balloon is rising at 5.0 m/s, this speed combines with the horizontal speed of the ball when calculating its initial velocity from the ground observer's perspective.
• Conversely, if it is descending, that 5.0 m/s subtracts from the vertical component of the overall initial velocity of the ball.

These dynamics show how vertical motion affects projectile analysis and why understanding the environment from which a projectile is released is essential.

Trigonometry in Physics

Trigonometry is a powerful tool in physics, especially for analyzing projectile motion. It helps to break down movements into vertical and horizontal components and find angles related to motion.

• Trigonometric functions like sine, cosine, and tangent allow you to determine the direction of a projectile's motion based on its components in different axes.
• In this exercise, we use the tangent function to find the angle made by the ball with the horizontal after it's thrown.

The angle, \( \theta \), calculated using \( \tan(\theta) = \frac{v_y}{v_x} \), allows us to see how the vertical and horizontal motions combine into a single vector. This helps describe the complete motion relative to an observer on the ground.
For instance:

• For the rising balloon scenario, the angle is approximately 26.6°, determining how much the ball leans upward as opposed to just moving horizontally.
• For the descending scenario, the angle is -26.6°, showing the ball's motion as it inclines downward.

Thus, trigonometry not only simplifies the process of analyzing projectile paths but also provides a clearer insight into how various components of motion work together.