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 the target must the rifle be aimed in order to hit the target?

The bullet is fired with an initial speed of 630 m/s towards a target 700 m away at the same level. Since the rifle is aimed directly at the center of the target, the initial velocity is horizontal. This means the initial vertical velocity is 0 m/s, and the initial horizontal velocity is 630 m/s.

To find how far above or below the center the bullet hits, we need to determine the vertical displacement when the bullet has traveled 700 m horizontally. The horizontal motion has no acceleration, so the time taken to cover 700 m is given by the horizontal distance divided by the horizontal velocity:

$$ t = \frac{\text{horizontal distance}}{\text{horizontal velocity}} = \frac{700}{630} $$

Simplifying the fraction:

$$ \frac{700}{630} = \frac{700 \div 70}{630 \div 70} = \frac{10}{9} \text{ seconds} $$

In the vertical direction, the initial velocity is 0 m/s, and acceleration due to gravity is $$ g = 9.8 \text{m/s}^2 $$ downward. The vertical displacement $$ h $$ is given by the equation of motion:

$$ h = u_y t + \frac{1}{2} a t^2 $$

Since $$ u_y = 0 $$ and $$ a = g $$ (taking downward as positive for displacement magnitude), this becomes:

$$ h = \frac{1}{2} g t^2 $$

Substituting $$ g = 9.8 $$ and $$ t = \frac{10}{9} $$:

$$ h = \frac{1}{2} \times 9.8 \times \left( \frac{10}{9} \right)^2 $$

First, compute $$ \left( \frac{10}{9} \right)^2 $$:

$$ \left( \frac{10}{9} \right)^2 = \frac{100}{81} $$

Now, multiply:

$$ h = \frac{1}{2} \times 9.8 \times \frac{100}{81} = \frac{1}{2} \times \frac{9.8 \times 100}{81} $$

Calculate $$ 9.8 \times 100 = 980 $$:

$$ h = \frac{1}{2} \times \frac{980}{81} = \frac{980}{2 \times 81} = \frac{980}{162} $$

Simplify by dividing numerator and denominator by 2:

$$ \frac{980 \div 2}{162 \div 2} = \frac{490}{81} $$

Now, compute the numerical value:

$$ \frac{490}{81} \approx 6.04938 \text{ m} $$

The negative sign indicates the displacement is downward, meaning the bullet hits below the center. The magnitude of the vertical displacement is approximately 6.05 m. Comparing with the options:

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

The value 6.05 m is closest to 6.1 m.

Hence, the correct answer is Option C.