A bomb at rest is exploded and the pieces are scattered in all directions with a maximum velocity of `20ms^(-1)`.Dangerous distance from that spot is `(g=10 m//s^(2))`

Options

A. `10 m`
B. `20 m`
C. `30 m`
D. `40 m`

Explanation

To solve the problem of determining the dangerous distance from the spot of the bomb explosion, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values**:
- Maximum velocity of the pieces, \( V = 20 \, \text{m/s} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

2. **Determine the Angle for Maximum Range**:
- The maximum range for a projectile is achieved at an angle of \( 45^\circ \).

3. **Use the Range Formula**:
- The formula for the range \( R \) of a projectile launched with an initial velocity \( V \) at an angle \( \theta \) is given by:
\[ R = \frac{V^2 \sin(2\theta)}{g} \]
- For \( \theta = 45^\circ \), \( \sin(90^\circ) = 1 \), so the formula simplifies to:
\[ R = \frac{V^2}{g} \]

4. **Substitute the Values**:
- Substitute \( V = 20 \, \text{m/s} \) and \( g = 10 \, \text{m/s}^2 \) into the formula:
\[ R = \frac{(20)^2}{10} \]

5. **Calculate the Range**:
- Calculate \( R \):
\[ R = \frac{400}{10} = 40 \, \text{m} \]

6. **Conclusion**:
- The dangerous distance from the spot of the bomb explosion is \( 40 \, \text{m} \).

### Final Answer:
The dangerous distance from the spot is **40 meters**.

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