If the man release the ball half meter above the hilltop, find the Potential Energy (PE ), Kinetic Energy (KE)  and Mechanical Energy (ME) of the  ball at Point A and Point B. Mass of ball=0.055 kg. What is the velocity of the ball at  point B?

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If the man release the ball half meter above the hilltop, find the Potential Energy (PE ), Kinetic Energy (KE)  and Mechanical Energy (ME) of the  ball at Point A and Point B. Mass of ball=0.055 kg. What is the velocity of the ball at  point B? 

### Physics of Free Fall

Understanding the concepts of free fall and projectile motion is essential in physics. Let’s explore an example using a simple illustration.

#### The Scenario

Imagine a person standing at the edge of a cliff, releasing a ball into the air. The ball is dropped from a height of 8.5 meters above the surface of the water. We will analyze its position at different points during the fall. The diagram provided illustrates the positions of the ball at three stages: 

- **Point A**: The initial position where the ball is released.
- **Point B**: Midway through the fall (mid-air).
- **Point C**: Just before making contact with the water's surface.

#### Diagram Explanation

1. **Initial Position (Point A)**:
   - At this point, the ball is at the edge of the cliff, 8.5 meters above the water. 
   - The ball starts with an initial velocity of 0 m/s (assuming it is simply dropped and not thrown).

2. **Mid-Air (Point B)**:
   - This point represents the ball during its descent. 
   - Here, the ball is accelerating due to gravity, which is approximately \( 9.8 m/s^2 \).

3. **Final Position (Point C)**:
   - This point represents the moment just before the ball hits the water's surface.
   - The velocity of the ball will be at its maximum just before impact due to the continuous acceleration from gravity through its descent.

#### Key Concepts

1. **Height (h)**: The vertical distance from the point of release to the surface of the water, which is 8.5 meters.
2. **Gravitational Acceleration (g)**: The constant acceleration due to gravity, approximately \( 9.8 m/s^2 \).

#### Equations of Motion

To calculate the position and velocity of the ball at any given time, we use the equations of motion under constant acceleration:

1. \( v = u + gt \)
   - \( v \) = final velocity
   - \( u \) = initial velocity (0 m/s)
   - \( g \) = acceleration due to gravity \( 9.8 m/s^2 \)
   - \( t \) = time

2. \( h = ut + \frac{1}{2}gt^2 \)
   - \( h \) = height

Using these
Transcribed Image Text:### Physics of Free Fall Understanding the concepts of free fall and projectile motion is essential in physics. Let’s explore an example using a simple illustration. #### The Scenario Imagine a person standing at the edge of a cliff, releasing a ball into the air. The ball is dropped from a height of 8.5 meters above the surface of the water. We will analyze its position at different points during the fall. The diagram provided illustrates the positions of the ball at three stages: - **Point A**: The initial position where the ball is released. - **Point B**: Midway through the fall (mid-air). - **Point C**: Just before making contact with the water's surface. #### Diagram Explanation 1. **Initial Position (Point A)**: - At this point, the ball is at the edge of the cliff, 8.5 meters above the water. - The ball starts with an initial velocity of 0 m/s (assuming it is simply dropped and not thrown). 2. **Mid-Air (Point B)**: - This point represents the ball during its descent. - Here, the ball is accelerating due to gravity, which is approximately \( 9.8 m/s^2 \). 3. **Final Position (Point C)**: - This point represents the moment just before the ball hits the water's surface. - The velocity of the ball will be at its maximum just before impact due to the continuous acceleration from gravity through its descent. #### Key Concepts 1. **Height (h)**: The vertical distance from the point of release to the surface of the water, which is 8.5 meters. 2. **Gravitational Acceleration (g)**: The constant acceleration due to gravity, approximately \( 9.8 m/s^2 \). #### Equations of Motion To calculate the position and velocity of the ball at any given time, we use the equations of motion under constant acceleration: 1. \( v = u + gt \) - \( v \) = final velocity - \( u \) = initial velocity (0 m/s) - \( g \) = acceleration due to gravity \( 9.8 m/s^2 \) - \( t \) = time 2. \( h = ut + \frac{1}{2}gt^2 \) - \( h \) = height Using these
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