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In the figure, a ball is thrown leftward from the left edge of the roof, at height h above the ground. The ball hits the ground 1.40 s later, at distance d = 27.0 m from the building and at angle θ = 66.0° with the horizontal. (a) Find h. (Hint: One way is to reverse the motion, as if on videotape.) What are the (b) magnitude and (c) angle relative to the horizontal of the velocity at which the ball is thrown (positive angle for above horizontal, negative for below)? I just need (c) solved. The answer is not 56.10 degrees.

In the figure, a ball is thrown leftward from the left edge of the roof, at height h above the ground. The ball hits the ground 1.40 s later, at distance d = 27.0 m from the building and at angle θ = 66.0° with the horizontal. (a) Find h. (Hint: One way is to reverse the motion, as if on videotape.) What are the (b) magnitude and (c) angle relative to the horizontal of the velocity at which the ball is thrown (positive angle for above horizontal, negative for below)? I just need (c) solved. The answer is not 56.10 degrees.

Principles of Physics: A Calculus-Based Text
Principles of Physics: A Calculus-Based Text
5th Edition
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter3: Motion In Two Dimensions
Section: Chapter Questions
Problem 3OQ: Figure OQ3.3 shows a birds-eye view of a car going around a highway curve. As the car moves from...
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In the figure, a ball is thrown leftward from the left edge of the roof, at height h above the ground. The ball hits the ground 1.40 s later, at distance d = 27.0 m from the building and at angle θ = 66.0° with the horizontal. (a) Find h. (Hint: One way is to reverse the motion, as if on videotape.) What are the (b) magnitude and (c) angle relative to the horizontal of the velocity at which the ball is thrown (positive angle for above horizontal, negative for below)?

I just need (c) solved. The answer is not 56.10 degrees. 

### Projectile Motion and Angle of Launch In this diagram, we observe a common scenario used to illustrate the principles of projectile motion in physics. Below we provide a detailed explanation to help you understand the key components and what they represent. #### Diagram Description: - **Building:** On the right side of the image, there is a depiction of a building with several windows. This building has a height represented by the variable \( h \). - **Ground:** The horizontal line at the bottom signifies the ground level on which both the building and the launching point are positioned. - **Projectile Launch Point:** On the left side, there is a launcher positioned at an incline. The angle of inclination is denoted by the variable \( \theta \). - **Angle of Projection (\(\theta\))**: This is the angle at which the object is launched relative to the horizontal ground. - **Distance (\(d\))**: The horizontal distance from the base of the launch point to the base of the building is represented by \( d \). #### Explanation of Components: 1. **Height (\(h\)):** - This is the vertical distance from the ground to the top of the building. It is an essential parameter when calculating the trajectory of a projectile aimed to reach the building's height. 2. **Horizontal Distance (\(d\)):** - This represents the horizontal separation from the launching point to the base of the building. It plays a crucial role in determining whether the projectile will hit the building. 3. **Angle of Launch (\(\theta\)):** - The angle at which the projectile is launched affects both the horizontal distance covered and the maximum height achieved by the projectile. Adjusting this angle will change the path the projectile takes. 4. **Path of Projectile:** - Though not explicitly drawn in this diagram, the projectile follows a curved path influenced by gravity which is typically parabolic in shape. The initial launch speed combined with the angle \( \theta \) determines the trajectory. ### Application in Calculations: When solving problems related to projectile motion with this setup, consider the following equations of motion: - **Horizontal Motion:** \( x = v_i \cos(\theta) \cdot t \) - **Vertical Motion:** \( y = v_i \sin(\theta) \cdot t - \frac{1}{2} g t^2 \) Where: - \( x \) and \( y \) are the horizontal and vertical
Transcribed Image Text:### Projectile Motion and Angle of Launch In this diagram, we observe a common scenario used to illustrate the principles of projectile motion in physics. Below we provide a detailed explanation to help you understand the key components and what they represent. #### Diagram Description: - **Building:** On the right side of the image, there is a depiction of a building with several windows. This building has a height represented by the variable \( h \). - **Ground:** The horizontal line at the bottom signifies the ground level on which both the building and the launching point are positioned. - **Projectile Launch Point:** On the left side, there is a launcher positioned at an incline. The angle of inclination is denoted by the variable \( \theta \). - **Angle of Projection (\(\theta\))**: This is the angle at which the object is launched relative to the horizontal ground. - **Distance (\(d\))**: The horizontal distance from the base of the launch point to the base of the building is represented by \( d \). #### Explanation of Components: 1. **Height (\(h\)):** - This is the vertical distance from the ground to the top of the building. It is an essential parameter when calculating the trajectory of a projectile aimed to reach the building's height. 2. **Horizontal Distance (\(d\)):** - This represents the horizontal separation from the launching point to the base of the building. It plays a crucial role in determining whether the projectile will hit the building. 3. **Angle of Launch (\(\theta\)):** - The angle at which the projectile is launched affects both the horizontal distance covered and the maximum height achieved by the projectile. Adjusting this angle will change the path the projectile takes. 4. **Path of Projectile:** - Though not explicitly drawn in this diagram, the projectile follows a curved path influenced by gravity which is typically parabolic in shape. The initial launch speed combined with the angle \( \theta \) determines the trajectory. ### Application in Calculations: When solving problems related to projectile motion with this setup, consider the following equations of motion: - **Horizontal Motion:** \( x = v_i \cos(\theta) \cdot t \) - **Vertical Motion:** \( y = v_i \sin(\theta) \cdot t - \frac{1}{2} g t^2 \) Where: - \( x \) and \( y \) are the horizontal and vertical
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