Problem 63 - Physics for Scientists and Engineers with Modern Physics

Question

A person in the passenger basket of a hot-air balloon throws a ball horizontally outward from the basket with speed \(10.0 \mathrm{~m} / \mathrm{s}\) (Fig. \(3-52\) ). 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: Understanding the Problem

We need to find the initial velocity of the ball relative to the ground observer. The given speeds are the horizontal speed of the ball from the basket and the vertical speed of the balloon. We will treat these velocities as vector components.

Step 2: Defining Relative Velocity Vectors

The horizontal speed of the ball rel. to the basket is \(v_{x} = 10.0 \, \text{m/s}\). For part (a), the balloon is rising; thus, its speed is upwards \(v_{y} = 5.0 \, \text{m/s}\). For part (b), it's descending, so \(v_{y} = -5.0 \, \text{m/s}\).

Step 3: Calculating the Magnitude of Initial Velocity (Part a)

For part (a), the initial velocity magnitude relative to the ground is given by \(v = \sqrt{v_{x}^2 + v_{y}^2}\). Substitute the values: \(v = \sqrt{10^2 + 5^2} = \sqrt{125} = 11.2 \, \text{m/s}\).

Step 4: Finding the Direction of Initial Velocity (Part a)

The direction \(\theta\) can be calculated using \(\tan \theta = \frac{v_{y}}{v_{x}}\). Thus, \(\theta = \tan^{-1}\left(\frac{5}{10}\right) = 26.6^\circ\) above the horizontal.

Step 5: Calculating the Magnitude of Initial Velocity (Part b)

For part (b), the initial velocity magnitude is \(v = \sqrt{v_{x}^2 + (-v_{y})^2}\). Substitute the values: \(v = \sqrt{10^2 + (-5)^2} = \sqrt{125} = 11.2 \, \text{m/s}\).

Step 6: Finding the Direction of Initial Velocity (Part b)

The direction \(\theta\) is calculated using \(\tan \theta = \frac{v_{y}}{v_{x}}\). Thus, \(\theta = \tan^{-1}\left(\frac{-5}{10}\right) = -26.6^\circ\), indicating the ball is 26.6° below the horizontal.

Key Concepts

Vector Components

When analyzing velocities, especially with an object in motion like our ball, it's essential to break them down into their **vector components**. These components often include a horizontal part and a vertical part.
For instance, in this exercise, the ball is given an initial horizontal speed of 10.0 m/s, which we can consider as the **x-component** of velocity: \(v_{x} = 10.0 \, \text{m/s}\). This is the speed at which the ball was thrown out of the basket.
Meanwhile, the hot-air balloon's motion affects the **y-component** of velocity, which is vertical. **In cases such as:**

• **Rising balloon**: The vertical speed \(v_{y}\) is 5.0 m/s upward.
• **Descending balloon**: The vertical speed \(v_{y}\) is -5.0 m/s downward.

These vector components allow us to form a complete picture of the ball's motion by combining these elements to calculate overall velocity. Each part plays a vital role in determining the magnitude and direction of the velocity vector.

Initial Velocity

The concept of **initial velocity** is critical here, as it represents the ball’s velocity relative to an observer on the ground at the moment the throw occurs.
To find this, we combine the horizontal and vertical vector components discussed earlier. This combination helps determine the initial velocity as a whole
Using **Pythagorean Theorem**, we can calculate the resultant velocity magnitude \(v\):

• In **part (a)**: Balloon rising, \(v = \sqrt{10^2 + 5^2} = \sqrt{125} = 11.2 \, \text{m/s}\).


• In **part (b)**: Balloon descending, \(v = \sqrt{10^2 + (-5)^2} = \sqrt{125} = 11.2 \, \text{m/s}\).

This shows that the magnitude of initial velocity remains consistent regardless of the balloon's direction of motion, emphasizing the influence of initial conditions in a relative velocity scenario.

Magnitude and Direction

Understanding the **magnitude and direction** of velocity is like finding the compass that guides us in understanding the ball’s path.
**Magnitude** provides the speed value of initial velocity, which indicates how fast the ball is moving at the outset.
As computed, the magnitude in both scenarios is 11.2 m/s. This does not change with the rising or descending condition because it’s calculated as the diagonal resultant of the perpendicular vector components.
**Direction** refers to the path that the velocity vector points toward:

• **Part (a):** When the balloon rises, the direction \(\theta\) is measured above the horizontal using \(\tan \theta = \frac{v_{y}}{v_{x}}\). Calculating gives us \(\theta = \tan^{-1}\left(\frac{5}{10}\right) = 26.6^\circ\).


• **Part (b):** When descending, \(\theta = \tan^{-1}\left(\frac{-5}{10}\right) = -26.6^\circ\), indicating it’s below the horizontal.

These directional angles tell us how the orientation changes based on the vertical motion of the balloon, confirming the relative nature of velocity as observed from a different frame of reference.