A bullet is fired from a cannon with velocity 500 m/s. If the angle of projection is `15^(@)` and `g = 10m//s^(2)`. Then the range is

Options

A. `25 xx 10^(3)m`
B. `12.5 xx 10^(3)m`
C. `50 xx 10^(2)m`
D. `25 xx 10^(2)m`

Answer

To find the range of a bullet fired from a cannon, we can use the formula for the range \( R \) in projectile motion:

\[ R = \frac{U^2 \sin(2\theta)}{g} \]

where:
- \( U \) is the initial velocity,
- \( \theta \) is the angle of projection,
- \( g \) is the acceleration due to gravity.

### Step-by-Step Solution:

1. **Identify the given values:**
- Initial velocity \( U = 500 \, \text{m/s} \)
- Angle of projection \( \theta = 15^\circ \)
- Acceleration due to gravity \( g = 10 \, \text{m/s}^2 \)

2. **Calculate \( \sin(2\theta) \):**
- First, calculate \( 2\theta \):
\[ 2\theta = 2 \times 15^\circ = 30^\circ \]
- Now, find \( \sin(30^\circ) \):
\[ \sin(30^\circ) = \frac{1}{2} \]

3. **Substitute the values into the range formula:**
\[ R = \frac{(500)^2 \cdot \sin(30^\circ)}{10} \]

4. **Calculate \( U^2 \):**
\[ U^2 = 500^2 = 250000 \]

5. **Substitute \( U^2 \) and \( \sin(30^\circ) \) into the equation:**
\[ R = \frac{250000 \cdot \frac{1}{2}}{10} \]

6. **Simplify the equation:**
\[ R = \frac{250000 \cdot 0.5}{10} = \frac{125000}{10} = 12500 \, \text{m} \]

7. **Final result:**
\[ R = 12500 \, \text{m} = 12.5 \times 10^3 \, \text{m} \]

### Conclusion:
The range of the bullet fired from the cannon is \( 12500 \, \text{m} \) or \( 12.5 \times 10^3 \, \text{m} \).