The gravitational acceleration of this planet is 9.8 m/s/s Find the horizontal and vertical speeds of the ball at the moment of the kick. Measure the horizontal displacement of the ball. Determine the time of flight for the ball. Determine the height of the ball when it hits the wall.
The gravitational acceleration of this planet is 9.8 m/s/s Find the horizontal and vertical speeds of the ball at the moment of the kick. Measure the horizontal displacement of the ball. Determine the time of flight for the ball. Determine the height of the ball when it hits the wall.
Physics for Scientists and Engineers: Foundations and Connections
1st Edition
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Katz, Debora M.
Chapter4: Two-and-three Dimensional Motion
Section: Chapter Questions
Problem 61PQ: You are watching a friend practice archery when he misses the target completely, and the arrow...
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Time of Flight (ms)=?
Height Hitting Wall (m)=?
Ground to Wall Soccer Kick Problem
Transcribed Image Text:### Projectile Motion Analysis
In this exercise, we will analyze the motion of a ball that is kicked, involving various calculations related to projectile motion. We will use the following information and carry out certain tasks.
#### Given Data:
- **Gravitational Acceleration**: 9.8 m/s²
#### Tasks to Complete:
1. **Find the horizontal and vertical speeds of the ball at the moment of the kick.**
- This step involves decomposing the initial velocity into horizontal and vertical components using the initial speed and launch angle.
2. **Measure the horizontal displacement of the ball.**
- Calculate the total distance the ball travels in the horizontal direction from the point of kick to the point it impacts the ground or another surface.
3. **Determine the time of flight for the ball.**
- Calculate the total duration the ball remains in the air, from the moment of the kick to when it lands.
4. **Determine the height of the ball when it hits the wall.**
- If a wall is present, calculate the height of the ball at the point where it strikes the wall.
Ensure to use the equations of motion and other relevant physics principles to solve these tasks accurately.
![### Projectile Motion Visualization
This diagram illustrates the trajectory of a projectile launched from the ground.
#### Graph Description
- **Background Grid**: The main background consists of a coordinate grid overlaid on a sky-blue gradient, which represents the trajectory plane.
- **Axes**:
- The **horizontal axis** (x-axis) at the bottom is marked in increments of 10 meters, extending up to 50 meters. Each major division corresponds to 10 meters.
- The **vertical axis** (y-axis) on the right side is similarly marked with the same 10-meter increments, extending up to 20 meters.
- **Ground**: The horizontal green strip at the bottom simulates the ground level.
- **Trajectory Line**: A dark red vertical line is drawn at the 10-meter mark to indicate a point of interest in the trajectory. The exact reason for this highlight is not specified.
#### Inset Detail
- **Projectile Launch Parameters**: The inset, located to the left of the main plot, offers a more detailed view of the launch conditions:
- **Velocity Vector**: An arrow indicating the launch velocity, marked as **21.9 m/s**, that represents the initial speed of the projectile.
- **Launch Angle**: The arrow emerges from a circle (representing the projectile) at an angle of **37.6°** above the horizontal.
This visual is useful for studying basic principles of projectile motion in physics, including:
- The parabolic path of a projectile.
- The effects of initial speed and launch angle on the range and height of the projectile.
For further examination, you can apply kinematic equations for projectile motion:
\[ y = x \tan(\theta) - \frac{gx^2}{2(v_0 \cos(\theta))^2} \]
Where:
- \( y \) is the vertical displacement.
- \( x \) is the horizontal displacement.
- \( \theta \) is the launch angle.
- \( g \) is the acceleration due to gravity (9.8 m/s²).
- \( v_0 \) is the initial velocity.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F54e8c682-be67-4ef0-bd1d-cf8b2e543b76%2F714c0310-cc9f-4a90-aa08-8262d2ac0e90%2Fskde715_processed.png&w=3840&q=75)
Transcribed Image Text:### Projectile Motion Visualization
This diagram illustrates the trajectory of a projectile launched from the ground.
#### Graph Description
- **Background Grid**: The main background consists of a coordinate grid overlaid on a sky-blue gradient, which represents the trajectory plane.
- **Axes**:
- The **horizontal axis** (x-axis) at the bottom is marked in increments of 10 meters, extending up to 50 meters. Each major division corresponds to 10 meters.
- The **vertical axis** (y-axis) on the right side is similarly marked with the same 10-meter increments, extending up to 20 meters.
- **Ground**: The horizontal green strip at the bottom simulates the ground level.
- **Trajectory Line**: A dark red vertical line is drawn at the 10-meter mark to indicate a point of interest in the trajectory. The exact reason for this highlight is not specified.
#### Inset Detail
- **Projectile Launch Parameters**: The inset, located to the left of the main plot, offers a more detailed view of the launch conditions:
- **Velocity Vector**: An arrow indicating the launch velocity, marked as **21.9 m/s**, that represents the initial speed of the projectile.
- **Launch Angle**: The arrow emerges from a circle (representing the projectile) at an angle of **37.6°** above the horizontal.
This visual is useful for studying basic principles of projectile motion in physics, including:
- The parabolic path of a projectile.
- The effects of initial speed and launch angle on the range and height of the projectile.
For further examination, you can apply kinematic equations for projectile motion:
\[ y = x \tan(\theta) - \frac{gx^2}{2(v_0 \cos(\theta))^2} \]
Where:
- \( y \) is the vertical displacement.
- \( x \) is the horizontal displacement.
- \( \theta \) is the launch angle.
- \( g \) is the acceleration due to gravity (9.8 m/s²).
- \( v_0 \) is the initial velocity.
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