A body falling freely under gravity passes two points 30 m apart in 1 s. From what point above the upper point it began to fall? (Take g = `9.8 m s^(-2)`).

Options

A. a. 32.1 m
B. b. 16.0 m
C. c. 8.6 m
D. d. 4.0 m

Explanation

To solve the problem, we need to determine the height from which the body fell above the upper point (point A) after passing two points that are 30 meters apart in 1 second. We will use the equations of motion under uniform acceleration due to gravity.

### Step-by-Step Solution:

1. **Identify the Known Values**:
- Distance between points A and B, \( s_{AB} = 30 \, m \)
- Time taken to travel from A to B, \( t_{AB} = 1 \, s \)
- Acceleration due to gravity, \( g = 9.8 \, m/s^2 \)

2. **Define Variables**:
- Let \( s_1 \) be the distance from the initial point to point A.
- Let \( s_2 \) be the distance from the initial point to point B.
- The relationship between \( s_1 \) and \( s_2 \) can be expressed as:
\[ s_2 - s_1 = 30 \, m \]

3. **Use the Equation of Motion**:
- For the distance \( s_1 \) (from the initial point to point A):
\[ s_1 = ut + \frac{1}{2} g t^2 \]
Since the body starts from rest, \( u = 0 \):
\[ s_1 = \frac{1}{2} g t^2 \]

- For the distance \( s_2 \) (from the initial point to point B), we need to consider the time taken to reach point B:
\[ s_2 = u(t + 1) + \frac{1}{2} g (t + 1)^2 \]
Again, since \( u = 0 \):
\[ s_2 = \frac{1}{2} g (t + 1)^2 \]

4. **Set Up the Equation**:
- Substitute \( s_1 \) and \( s_2 \) into the distance equation:
\[ s_2 - s_1 = 30 \]
This gives:
\[ \frac{1}{2} g (t + 1)^2 - \frac{1}{2} g t^2 = 30 \]

5. **Simplify the Equation**:
- Factor out \( \frac{1}{2} g \):
\[ \frac{1}{2} g \left( (t + 1)^2 - t^2 \right) = 30 \]
- Expand \( (t + 1)^2 \):
\[ (t^2 + 2t + 1) - t^2 = 2t + 1 \]
- Thus, we have:
\[ \frac{1}{2} g (2t + 1) = 30 \]

6. **Solve for \( t \)**:
- Substitute \( g = 9.8 \):
\[ \frac{1}{2} \cdot 9.8 (2t + 1) = 30 \]
\[ 4.9 (2t + 1) = 30 \]
\[ 2t + 1 = \frac{30}{4.9} \approx 6.12 \]
\[ 2t = 6.12 - 1 \approx 5.12 \]
\[ t \approx 2.56 \, s \]

7. **Calculate \( s_1 \)**:
- Now substitute \( t \) back into the equation for \( s_1 \):
\[ s_1 = \frac{1}{2} g t^2 = \frac{1}{2} \cdot 9.8 \cdot (2.56)^2 \]
\[ s_1 \approx \frac{1}{2} \cdot 9.8 \cdot 6.5536 \approx 32.1 \, m \]

8. **Final Answer**:
- The height from which the body began to fall above point A is approximately **32.1 meters**.