A projectile is fired at an upward angle of 45.6° from the top of a 342m cliff with a speed of 153m/s. What will be its speed when it strikes the ground below?

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Chapter1: Units, Trigonometry. And Vectors
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### Projectile Motion Problem

**Problem Statement:**

A projectile is fired at an upward angle of 45.6° from the top of a 342m cliff with a speed of 153 m/s. What will be its speed when it strikes the ground below?

**Solution Approach:**

To solve this problem, we will use principles from projectile motion and kinematics. The problem can be broken down into the following steps:

1. **Initial Velocity Components:**
   Calculate the horizontal and vertical components of the initial velocity.
   
   - \( V_{ix} = V_i \cos(\theta) \)
   - \( V_{iy} = V_i \sin(\theta) \)
   
   Where \( V_i = 153 \, \text{m/s} \) and \( \theta = 45.6° \).

2. **Time of Flight:**
   Use the vertical motion equation to find the time it takes for the projectile to hit the ground.
   
   - Considering the equation \( y = V_{iy} t + \frac{1}{2} g t^2 \), solve for \( t \) given \( y = -342 \, \text{m} \) (downward displacement from the cliff).

3. **Final Velocity Components:**
   Determine the final vertical velocity \( V_{fy} \) using the equation \( V_{fy} = V_{iy} + g t \), where \( g = 9.81 \, \text{m/s}^2 \).

4. **Resultant Velocity:**
   Calculate the magnitude of the resultant velocity \( V_f \) when the projectile hits the ground using the Pythagorean theorem:
   
   - \( V_f = \sqrt{V_{fx}^2 + V_{fy}^2} \)
   
   Where \( V_{fx} \) remains the same as \( V_{ix} \), given there is no horizontal acceleration in ideal projectile motion.

**Diagram Explanation:**

If a diagram were included, it would show:
1. A projectile being fired from the top of a cliff.
2. The cliff’s height labeled as 342m.
3. The angle of projection, 45.6°, marked from the horizontal.
4. Initial velocity vector split into horizontal (\( V_{ix} \)) and vertical (\( V_{iy} \)) components.
5. The parabolic trajectory
Transcribed Image Text:### Projectile Motion Problem **Problem Statement:** A projectile is fired at an upward angle of 45.6° from the top of a 342m cliff with a speed of 153 m/s. What will be its speed when it strikes the ground below? **Solution Approach:** To solve this problem, we will use principles from projectile motion and kinematics. The problem can be broken down into the following steps: 1. **Initial Velocity Components:** Calculate the horizontal and vertical components of the initial velocity. - \( V_{ix} = V_i \cos(\theta) \) - \( V_{iy} = V_i \sin(\theta) \) Where \( V_i = 153 \, \text{m/s} \) and \( \theta = 45.6° \). 2. **Time of Flight:** Use the vertical motion equation to find the time it takes for the projectile to hit the ground. - Considering the equation \( y = V_{iy} t + \frac{1}{2} g t^2 \), solve for \( t \) given \( y = -342 \, \text{m} \) (downward displacement from the cliff). 3. **Final Velocity Components:** Determine the final vertical velocity \( V_{fy} \) using the equation \( V_{fy} = V_{iy} + g t \), where \( g = 9.81 \, \text{m/s}^2 \). 4. **Resultant Velocity:** Calculate the magnitude of the resultant velocity \( V_f \) when the projectile hits the ground using the Pythagorean theorem: - \( V_f = \sqrt{V_{fx}^2 + V_{fy}^2} \) Where \( V_{fx} \) remains the same as \( V_{ix} \), given there is no horizontal acceleration in ideal projectile motion. **Diagram Explanation:** If a diagram were included, it would show: 1. A projectile being fired from the top of a cliff. 2. The cliff’s height labeled as 342m. 3. The angle of projection, 45.6°, marked from the horizontal. 4. Initial velocity vector split into horizontal (\( V_{ix} \)) and vertical (\( V_{iy} \)) components. 5. The parabolic trajectory
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