An NCC parade is going at a uniform speed of 6 km/h through a place under a berry tree on which a bird is sitting at a height of 12.1 m. At a particular instant the bird drops a berry. Which cadet (ie the distance from the tree at the instant) will receive the berry on his uniform?

Explanation

To solve the problem step by step, we need to determine how far the cadet will be from the tree when the berry dropped by the bird reaches the ground.

### Step-by-Step Solution:

1. **Identify the height from which the berry is dropped**:
The bird is sitting at a height of 12.1 meters.

2. **Calculate the time taken for the berry to fall**:
We will use the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
Here,
- \(s = 12.1 \, \text{m}\) (distance fallen),
- \(u = 0 \, \text{m/s}\) (initial velocity, since the berry is dropped),
- \(a = 9.8 \, \text{m/s}^2\) (acceleration due to gravity).

Plugging in the values:
\[ 12.1 = 0 \cdot t + \frac{1}{2} \cdot 9.8 \cdot t^2 \]
Simplifying gives:
\[ 12.1 = 4.9 t^2 \]
\[ t^2 = \frac{12.1}{4.9} \approx 2.46 \]
\[ t \approx \sqrt{2.46} \approx 1.57 \, \text{s} \]

3. **Convert the speed of the parade from km/h to m/s**:
The speed of the parade is given as 6 km/h. We convert this to meters per second using the conversion factor \( \frac{5}{18} \):
\[ \text{Speed} = 6 \times \frac{5}{18} = \frac{30}{18} = \frac{5}{3} \, \text{m/s} \]

4. **Calculate the distance traveled by the parade during the time the berry falls**:
Using the formula for distance:
\[ \text{Distance} = \text{Speed} \times \text{Time} \]
We have:
\[ \text{Distance} = \frac{5}{3} \times 1.57 \approx \frac{5 \times 1.57}{3} \approx 2.62 \, \text{m} \]

5. **Conclusion**:
The cadet will be approximately 2.62 meters away from the tree when the berry reaches the ground.