A rock is projected from the edge of the top of a building with an initial velocity of 12.2 m/s at an angle of 53° above the horizontal. The rock strikes the ground a horizontal distance of 25 m from the base of the building. Assume that the ground is level and that the side of the building is vertical. How tall is the building?

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

#### Problem Statement:
A rock is projected from the edge of the top of a building with an initial velocity of 12.2 m/s at an angle of 53° above the horizontal. The rock strikes the ground a horizontal distance of 25 meters from the base of the building. Assuming that the ground is level and that the side of the building is vertical, determine the height of the building.

#### Explanation:

1. **Given Data:**
   - Initial velocity (\(v_0\)) = 12.2 m/s
   - Angle of projection (\(\theta\)) = 53°
   - Horizontal distance traveled (Range, \(R\)) = 25 meters

2. **Components of Initial Velocity:**
   - Horizontal component (\(v_{0x}\)): \(v_{0x} = v_0 \cdot \cos(\theta) = 12.2 \cdot \cos(53°)\)
   - Vertical component (\(v_{0y}\)): \(v_{0y} = v_0 \cdot \sin(\theta) = 12.2 \cdot \sin(53°)\)

3. **Time of Flight:**
   - The time of flight (\(t\)) for horizontal motion can be calculated using the range and horizontal component of the initial velocity:
     \[
     t = \frac{R}{v_{0x}} = \frac{25}{12.2 \cdot \cos(53°)}
     \]

4. **Vertical Motion:**
   - The height (h) of the building can be determined using the vertical motion equation:
     \[
     h = v_{0y} \cdot t + \frac{1}{2} \cdot (-g) \cdot t^2
     \]
     Where \(g\) is the acceleration due to gravity (9.8 m/s²).

5. **Calculation Steps:**
   - Calculate the horizontal (\(v_{0x}\)) and vertical (\(v_{0y}\)) components of the initial velocity.
   - Determine the time of flight (\(t\)).
   - Use the time of flight to calculate the height (h) of the building.
    
This problem involves applying concepts of projectile motion, breaking down the motion into horizontal and vertical components, and using kinematic equations to find the height
Transcribed Image Text:### Physics Problem: Projectile Motion #### Problem Statement: A rock is projected from the edge of the top of a building with an initial velocity of 12.2 m/s at an angle of 53° above the horizontal. The rock strikes the ground a horizontal distance of 25 meters from the base of the building. Assuming that the ground is level and that the side of the building is vertical, determine the height of the building. #### Explanation: 1. **Given Data:** - Initial velocity (\(v_0\)) = 12.2 m/s - Angle of projection (\(\theta\)) = 53° - Horizontal distance traveled (Range, \(R\)) = 25 meters 2. **Components of Initial Velocity:** - Horizontal component (\(v_{0x}\)): \(v_{0x} = v_0 \cdot \cos(\theta) = 12.2 \cdot \cos(53°)\) - Vertical component (\(v_{0y}\)): \(v_{0y} = v_0 \cdot \sin(\theta) = 12.2 \cdot \sin(53°)\) 3. **Time of Flight:** - The time of flight (\(t\)) for horizontal motion can be calculated using the range and horizontal component of the initial velocity: \[ t = \frac{R}{v_{0x}} = \frac{25}{12.2 \cdot \cos(53°)} \] 4. **Vertical Motion:** - The height (h) of the building can be determined using the vertical motion equation: \[ h = v_{0y} \cdot t + \frac{1}{2} \cdot (-g) \cdot t^2 \] Where \(g\) is the acceleration due to gravity (9.8 m/s²). 5. **Calculation Steps:** - Calculate the horizontal (\(v_{0x}\)) and vertical (\(v_{0y}\)) components of the initial velocity. - Determine the time of flight (\(t\)). - Use the time of flight to calculate the height (h) of the building. This problem involves applying concepts of projectile motion, breaking down the motion into horizontal and vertical components, and using kinematic equations to find the height
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