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?

To solve for the height of the building, we need to break the motion of the rock into its horizontal and vertical components using basic projectile motion equations. Here's how we can approach the problem:

### 1. **Resolve the initial velocity into components:**

The initial velocity is given as 12.2 m/s at an angle of 53° above the horizontal. The horizontal and vertical components of the velocity can be found using trigonometry:

- Horizontal velocity component:  
  \( v_{x0} = v_0 \cos(\theta) = 12.2 \cos(53^\circ) \)
  
- Vertical velocity component:  
  \( v_{y0} = v_0 \sin(\theta) = 12.2 \sin(53^\circ) \)

### 2. **Horizontal motion:**

The horizontal motion is uniform, meaning the horizontal velocity remains constant throughout the motion. The horizontal distance is given as 25 m, so we can use the following formula to find the time of flight:

\[
x = v_{x0} t
\]

Where:
- \( x = 25 \, \text{m} \) is the horizontal distance.
- \( v_{x0} \) is the horizontal component of the velocity.
- \( t \) is the time of flight.

Rearranging for \( t \):

\[
t = \frac{x}{v_{x0}}
\]

### 3. **Vertical motion:**

For the vertical motion, we can use the kinematic equation to find the height of the building. The vertical displacement from the top of the building to the ground is \( -h \), where \( h \) is the height of the building. The vertical displacement is given by:

\[
y = v_{y0} t - \frac{1}{2} g t^2
\]

Where:
- \( y = -h \) (since the rock lands on the ground, it’s a downward displacement),
- \( v_{y0} \) is the initial vertical velocity,
- \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity.

Rearranging for \( h \):

\[
h = v_{y0} t - \frac{1}{2} g t^2
\]

Now let's compute each part step by step to find the height of the building.The height of the building is approximately \( 23.63 \, \text{m} \). The negative sign indicates that the rock fell from that height to the ground level.

Would you like further details or clarifications?

Here are 5 related questions to explore:
1. How would the time of flight change if the initial velocity were increased?
2. What would happen if the angle of projection was smaller?
3. How can air resistance affect the result?
4. How would the problem change if the ground was inclined instead of level?
5. What if the building's height were known and the horizontal distance needed to be calculated?

**Tip:** In projectile motion, the horizontal and vertical components are independent of each other, but they share the same time of flight.

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?

Learn how to solve a projectile motion problem where a rock is projected with an initial velocity of 12.2 m/s at a 53° angle. This problem includes key concepts in kinematics, such as horizontal and vertical motion, and uses trigonometry to find the height of the building. Suitable for students in Grades 10-12.

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