A body falling freely under the action of gravity passes 2 points 9 m apart vertically in 0.2 s. From what height above the higher point did it start to fall?

Options

A. 99 m
B. 200 m
C. 20 m
D. 109 m

Explanation

To solve the problem, we need to determine the height from which the body started to fall above the higher point (point A) when it falls freely under the influence of gravity.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have two points, A and B, which are 9 meters apart vertically.
- The time taken to fall between these two points is 0.2 seconds.
- We need to find the height (h) from which the body started falling above point A.

2. **Setting Up the Equations**:
- Let the height from which the body falls be h (above point A).
- The distance fallen from point C (starting point) to point A is h.
- The distance fallen from point A to point B is 9 m.

3. **Using the Equation of Motion**:
- For the fall from point C to point A:
\[ h = ut + \frac{1}{2} g t^2 \]
Here, \( u = 0 \) (initial velocity), \( g = 9.8 \, \text{m/s}^2 \), and \( t \) is the time taken to fall from C to A.

- Therefore, the equation simplifies to:
\[ h = \frac{1}{2} g t^2 \]

4. **Calculating the Distance from A to B**:
- For the fall from point A to point B:
\[ 9 = u' t' + \frac{1}{2} g t'^2 \]
Here, \( u' \) is the velocity at point A, and \( t' = 0.2 \, \text{s} \).

5. **Finding the Velocity at Point A**:
- The velocity at point A can be calculated using:
\[ v = u + gt \]
Since the body is falling from point C to A, we can substitute \( u = 0 \):
\[ v = 0 + g t \]

6. **Using the Velocity to Find Distance from A to B**:
- Substitute \( v \) into the equation for the distance from A to B:
\[ 9 = (g t) t' + \frac{1}{2} g (t')^2 \]
- Substitute \( g = 9.8 \, \text{m/s}^2 \) and \( t' = 0.2 \):
\[ 9 = (9.8 t) (0.2) + \frac{1}{2} (9.8) (0.2)^2 \]

7. **Solving for t**:
- Simplifying the equation:
\[ 9 = 1.96 t + 0.196 \]
- Rearranging gives:
\[ 9 - 0.196 = 1.96 t \]
\[ 8.804 = 1.96 t \]
\[ t = \frac{8.804}{1.96} \approx 4.49 \, \text{s} \]

8. **Finding h**:
- Now substitute \( t \) back into the equation for h:
\[ h = \frac{1}{2} g t^2 \]
\[ h = \frac{1}{2} (9.8) (4.49)^2 \]
\[ h \approx \frac{1}{2} (9.8) (20.1) \approx 98.5 \, \text{m} \]

### Final Answer:
The height from which the body started to fall above point A is approximately **98.5 meters**.