In the case of a body freely falling from small height

Options

A. the changes of position are equal in equal intervals of time
B. the changes of velocity are equal in unequal intervals of time
C. the changes of acceleration is zero in equal or unequal intervals of time
D. None

Answer

To solve the problem of a body freely falling from a small height, we will analyze the motion step by step.

### Step-by-Step Solution:

1. **Understanding the Context**:
- We are considering a body falling freely from a small height \( h \) above the Earth's surface.
- The radius of the Earth is denoted as \( R \), and we assume \( h \) is much smaller than \( R \).

**Hint**: Remember that when dealing with small heights compared to the Earth's radius, we can approximate gravitational acceleration \( g \) as constant.

2. **Gravitational Acceleration**:
- The gravitational acceleration at height \( h \) is given by the formula:
\[ g' = g \left(1 - \frac{2h}{R}\right) \]
- Since \( h \) is much smaller than \( R \), we can approximate:
\[ g' \approx g \]
- This means that for small heights, the value of \( g' \) remains approximately equal to \( g \).

**Hint**: When \( h \) is very small compared to \( R \), the change in gravitational acceleration can be neglected.

3. **Initial Conditions**:
- When the body is released, its initial velocity \( u = 0 \).
- The acceleration \( a \) during free fall is equal to \( g \).

**Hint**: Always define your initial conditions clearly before proceeding with the equations of motion.

4. **Using the Equations of Motion**:
- The equation for distance \( h \) fallen in time \( t \) is given by:
\[ h = ut + \frac{1}{2} g t^2 \]
- Substituting \( u = 0 \):
\[ h = \frac{1}{2} g t^2 \]

**Hint**: This equation shows that the distance fallen is proportional to the square of the time, indicating accelerated motion.

5. **Relationship Between Height and Time**:
- Rearranging the equation gives:
\[ t^2 = \frac{2h}{g} \]
- This indicates that as time increases, the height \( h \) increases quadratically.

**Hint**: A quadratic relationship means that the body does not cover equal distances in equal intervals of time.

6. **Velocity Calculation**:
- The final velocity \( v \) at any time \( t \) can be calculated using:
\[ v = u + gt \]
- Since \( u = 0 \):
\[ v = gt \]

**Hint**: The velocity increases linearly with time due to constant acceleration.

7. **Conclusion on Motion**:
- The body does not cover equal distances in equal intervals of time, as the distance increases quadratically with time.
- The change in velocity is not constant; it increases linearly with time.

**Hint**: Recognize that in uniformly accelerated motion, the distance covered increases at a rate that depends on the square of time, while velocity increases linearly.

### Final Answer:
The change of position is unequal in equal intervals of time, and the change of velocity is equal in unequal intervals of time.