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Step-by-Step Solution
Step 1: Recall the formula for maximum range
When a projectile is launched with speed $v$ from the ground on level terrain (neglecting air resistance), the maximum horizontal range $R_{\text{max}}$ occurs at a projection angle of $45^\circ$, and is given by
$$
R_{\text{max}} = \frac{v^2}{g}.
$$
Step 2: Understand the area covered on the ground
If the projectile is fired in all possible directions from a point, the set of all possible landing points forms a circle on the ground, whose radius is $R_{\text{max}}$. Hence, the area $A$ covered by the projectile is
$$
A = \pi \left(R_{\text{max}}\right)^2 = \pi \left(\frac{v^2}{g}\right)^2
= \frac{\pi \, v^4}{g^2}.
$$
Step 3: Compute the area ratio for the two guns
Let the speeds of the bullets from guns A and B be $v_A$ and $v_B$ respectively.
• For Gun A: $v_A = 1\,\text{km/s}$.
The area covered by Gun A is
$$
A_A = \frac{\pi \, v_A^4}{g^2} = \frac{\pi \, (1)^4}{g^2} = \frac{\pi}{g^2}.
$$
• For Gun B: $v_B = 2\,\text{km/s}$.
The area covered by Gun B is
$$
A_B = \frac{\pi \, v_B^4}{g^2} = \frac{\pi \, (2)^4}{g^2} = \frac{\pi \, (16)}{g^2}.
$$
Step 4: Determine the ratio of the areas
Taking the ratio of the areas covered by guns A and B,
$$
\frac{A_A}{A_B}
\;=\; \frac{\frac{\pi}{g^2}}{\frac{\pi \cdot 16}{g^2}}
\;=\; \frac{1}{16}.
$$
Hence, the ratio is
$$
1 : 16.
$$
Answer: 1 : 16
