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. 32.1 m
B. 16.0 m
C. 8.6 m
D. 4.0 m

Explanation

To solve the problem step by step, we will follow the reasoning laid out in the video transcript.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have a body falling freely under gravity that passes two points (let's call them A and B) which are 30 meters apart in a time interval of 1 second. We need to find the height from which the body started falling above point A.

2. **Setting Up the Variables**:
- Let the time taken to reach point A be \( T \).
- Therefore, the time taken to reach point B will be \( T + 1 \) seconds.
- The distance between points A and B is given as 30 m.

3. **Using the Equations of Motion**:
The distance fallen from the starting point to point B can be expressed using the equation of motion:
\[ S_B = \frac{1}{2} g (T + 1)^2 \]
The distance fallen from the starting point to point A is:
\[ S_A = \frac{1}{2} g T^2 \]
The difference in distances \( S_B - S_A \) is equal to 30 m:
\[ S_B - S_A = 30 \]

4. **Substituting the Equations**:
Plugging in the equations for \( S_B \) and \( S_A \):
\[ \frac{1}{2} g (T + 1)^2 - \frac{1}{2} g T^2 = 30 \]
Factoring out \( \frac{1}{2} g \):
\[ \frac{1}{2} g \left((T + 1)^2 - T^2\right) = 30 \]

5. **Expanding the Equation**:
Expanding \( (T + 1)^2 \):
\[ (T + 1)^2 = T^2 + 2T + 1 \]
Therefore:
\[ (T + 1)^2 - T^2 = 2T + 1 \]
Substituting this back into the equation gives:
\[ \frac{1}{2} g (2T + 1) = 30 \]

6. **Solving for T**:
Rearranging the equation:
\[ g (2T + 1) = 60 \]
Using \( g = 9.8 \, \text{m/s}^2 \):
\[ 9.8 (2T + 1) = 60 \]
Dividing both sides by 9.8:
\[ 2T + 1 = \frac{60}{9.8} \approx 6.12 \]
Solving for \( T \):
\[ 2T = 6.12 - 1 \approx 5.12 \]
\[ T \approx 2.56 \, \text{s} \]

7. **Finding the Distance from the Starting Point to A**:
Now we need to find the distance \( S_A \) from the starting point to point A using:
\[ S_A = \frac{1}{2} g T^2 \]
Substituting \( T \) and \( g \):
\[ S_A = \frac{1}{2} \times 9.8 \times (2.56)^2 \]
Calculating \( (2.56)^2 \approx 6.5536 \):
\[ S_A \approx \frac{1}{2} \times 9.8 \times 6.5536 \approx 32.1 \, \text{m} \]

8. **Conclusion**:
The body began to fall from a height of approximately **32.1 meters** above point A.