The initial speed of a bullet fired from a rifle is 630 m/s. . The rifle is fired at the centre of a target 700 m away at the same level as the target. How far above the centre of target, the rifle must be aimed in order to hit the target?

Options

A. 6.1 m
B. 1.0 m
C. 9.8 m
D. 4.2m

Explanation

To solve the problem, we need to determine how far above the center of the target the rifle must be aimed in order to hit the target. The bullet is fired with an initial speed of 630 m/s at a target that is 700 m away.

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Initial speed of the bullet, \( u = 630 \, \text{m/s} \)
- Distance to the target, \( R = 700 \, \text{m} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \) (approximate value)

2. **Use the Range Formula**:
The range \( R \) of a projectile is given by the formula:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
Rearranging this formula to find \( \sin(2\theta) \):
\[ \sin(2\theta) = \frac{R \cdot g}{u^2} \]

3. **Substitute the Known Values**:
Substitute \( R = 700 \, \text{m} \), \( g = 10 \, \text{m/s}^2 \), and \( u = 630 \, \text{m/s} \) into the equation:
\[ \sin(2\theta) = \frac{700 \cdot 10}{630^2} \]
Calculate \( 630^2 \):
\[ 630^2 = 396900 \]
Now substitute:
\[ \sin(2\theta) = \frac{7000}{396900} \approx 0.0176 \]

4. **Calculate \( 2\theta \)**:
To find \( 2\theta \), take the inverse sine:
\[ 2\theta = \arcsin(0.0176) \]
Using a calculator, we find:
\[ 2\theta \approx 1.01^\circ \]
Therefore, \( \theta \approx 0.505^\circ \).

5. **Calculate the Height \( H \)**:
We know that:
\[ \tan(\theta) = \frac{H}{R} \]
Rearranging gives:
\[ H = R \cdot \tan(\theta) \]
Using \( R = 700 \, \text{m} \) and \( \tan(\theta) \approx \sin(\theta) \) for small angles:
\[ \tan(0.505^\circ) \approx 0.0088 \]
Thus:
\[ H = 700 \cdot 0.0088 \approx 6.16 \, \text{m} \]

6. **Final Result**:
The rifle must be aimed approximately \( H \approx 6.16 \, \text{m} \) above the center of the target to hit it.

### Summary:
The rifle must be aimed approximately **6.16 meters** above the center of the target.