A student standing on a cliff throws a rock from a vertical height of d=8.0md=8.0m above the level ground with velocity v0=16m/sv0=16m/s at an angle θ=26∘θ=26∘ below the horizontal, as shown. It moves without air resistance. Use a Cartesian coordinate system with the origin at the initial position of the rock. a. With what speed, in meters per second, does the stone strike the ground? b. If the rock had been thrown from the clifftop with the same initial speed and the same angle, but above the horizontal, would its impact velocity be different?
A student standing on a cliff throws a rock from a vertical height of d=8.0md=8.0m above the level ground with velocity v0=16m/sv0=16m/s at an angle θ=26∘θ=26∘ below the horizontal, as shown. It moves without air resistance. Use a Cartesian coordinate system with the origin at the initial position of the rock. a. With what speed, in meters per second, does the stone strike the ground? b. If the rock had been thrown from the clifftop with the same initial speed and the same angle, but above the horizontal, would its impact velocity be different?
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ISBN:9781305952300
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Chapter1: Units, Trigonometry. And Vectors
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Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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A student standing on a cliff throws a rock from a vertical height of d=8.0md=8.0m above the level ground with velocity v0=16m/sv0=16m/s at an angle θ=26∘θ=26∘ below the horizontal, as shown. It moves without air resistance. Use a Cartesian coordinate system with the origin at the initial position of the rock.
a. With what speed, in meters per second, does the stone strike the ground?
b. If the rock had been thrown from the clifftop with the same initial speed and the same angle, but above the horizontal, would its impact velocity be different?
![### Projectile Motion off a Cliff
The diagram represents a scenario involving projectile motion where an object is launched from the edge of a cliff. This scenario can be analyzed to understand the principles of two-dimensional kinematics.
#### Explanatory Components of the Diagram:
1. **Initial Position (Cliff Edge)**:
- The object is launched from the edge of a cliff at a certain height, \(d\).
2. **Coordinate System**:
- The origin of the coordinate system is located at the point of launch.
- The horizontal axis is labeled as \(x\).
- The vertical axis is labeled as \(y\).
3. **Initial Velocity (\(v_0\))**:
- The initial velocity (\(v_0\)) is depicted as a red arrow starting from the launch point, indicating both direction and magnitude.
- The initial velocity makes an angle \(\theta\) with the horizontal axis (\(x\)).
4. **Trajectory of the Object**:
- The dashed line indicates the parabolic trajectory of the projectile. This curve shows the path the object follows under the influence of gravity alone, neglecting air resistance.
5. **Height (\(d\))**:
- The vertical distance from the launch point to the ground, labeled \(d\), indicates the height of the cliff.
### Key Concepts:
1. **Kinematic Equations**:
The motion can be analyzed using the kinematic equations for horizontal and vertical components:
- Horizontal motion:
\[
x = v_0 \cos(\theta) \cdot t
\]
- Vertical motion:
\[
y = d + v_0 \sin(\theta) \cdot t - \frac{1}{2} g t^2
\]
Where:
- \(g\) is the acceleration due to gravity (approximately \(9.8 \, m/s^2\) on Earth's surface).
- \(t\) is the time since the projectile was launched.
2. **Range of the Projectile**:
The horizontal distance the projectile travels before hitting the ground can be termed as the range. This can be calculated by determining the total time of flight and substituting it back into the horizontal motion equation.
3. **Time of Flight**:
The time taken by the projectile to reach the ground can be found by solving the vertical](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Faf2a9245-87db-4e54-b147-768ccba693e7%2F26961e0a-6b22-4f70-a6e0-6feac7c068fe%2Fslyxmg_processed.png&w=3840&q=75)
Transcribed Image Text:### Projectile Motion off a Cliff
The diagram represents a scenario involving projectile motion where an object is launched from the edge of a cliff. This scenario can be analyzed to understand the principles of two-dimensional kinematics.
#### Explanatory Components of the Diagram:
1. **Initial Position (Cliff Edge)**:
- The object is launched from the edge of a cliff at a certain height, \(d\).
2. **Coordinate System**:
- The origin of the coordinate system is located at the point of launch.
- The horizontal axis is labeled as \(x\).
- The vertical axis is labeled as \(y\).
3. **Initial Velocity (\(v_0\))**:
- The initial velocity (\(v_0\)) is depicted as a red arrow starting from the launch point, indicating both direction and magnitude.
- The initial velocity makes an angle \(\theta\) with the horizontal axis (\(x\)).
4. **Trajectory of the Object**:
- The dashed line indicates the parabolic trajectory of the projectile. This curve shows the path the object follows under the influence of gravity alone, neglecting air resistance.
5. **Height (\(d\))**:
- The vertical distance from the launch point to the ground, labeled \(d\), indicates the height of the cliff.
### Key Concepts:
1. **Kinematic Equations**:
The motion can be analyzed using the kinematic equations for horizontal and vertical components:
- Horizontal motion:
\[
x = v_0 \cos(\theta) \cdot t
\]
- Vertical motion:
\[
y = d + v_0 \sin(\theta) \cdot t - \frac{1}{2} g t^2
\]
Where:
- \(g\) is the acceleration due to gravity (approximately \(9.8 \, m/s^2\) on Earth's surface).
- \(t\) is the time since the projectile was launched.
2. **Range of the Projectile**:
The horizontal distance the projectile travels before hitting the ground can be termed as the range. This can be calculated by determining the total time of flight and substituting it back into the horizontal motion equation.
3. **Time of Flight**:
The time taken by the projectile to reach the ground can be found by solving the vertical
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