Two guns A and B can fire bullets at speeds 1 km/s and 2 km/s respectively. From a point on a horizontal ground, they are fired in all possible directions. The ratio of maximum areas covered by the bullets fired by the two guns, on the ground is:

Let us consider one of the guns first. A bullet is projected from ground level with an initial speed $$v$$ and makes an angle $$\theta$$ with the horizontal.

For a projectile that starts and finishes at the same height, the horizontal range is given by the well-known kinematic result

$$R \;=\; \frac{v^{2}\sin 2\theta}{g},$$

where $$g$$ denotes the magnitude of gravitational acceleration.

We are asked for the maximum horizontal distance the bullet can reach. The factor $$\sin 2\theta$$ attains its greatest value, namely $$1$$, when $$2\theta = 90^{\circ}$$, i.e. $$\theta = 45^{\circ}$$. Hence the greatest possible range for speed $$v$$ is

$$R_{\max} \;=\; \frac{v^{2}}{g}.$$

The bullet can be fired in any azimuthal direction. Therefore every direction in the horizontal plane is available, and the set of all landing points forms a full circle centred on the firing point with radius $$R_{\max}$$.

The area of this circle is thus

$$A \;=\; \pi R_{\max}^{2} \;=\; \pi \left( \frac{v^{2}}{g} \right)^{2} \;=\; \pi \frac{v^{4}}{g^{2}}.$$

Observe that $$\pi$$ and $$g$$ are the same for both guns, so the area is proportional to the fourth power of the muzzle speed:

$$A \;\propto\; v^{4}.$$

Now compare the two specific guns.

For gun A we have $$v_{A}=1\ \text{km/s},$$ so $$A_{A}\propto (1)^{4}=1.$$

For gun B we have $$v_{B}=2\ \text{km/s},$$ so $$A_{B}\propto (2)^{4}=16.$$

Taking the ratio of the two areas,

$$\frac{A_{A}}{A_{B}} = \frac{1}{16}.$$

Therefore the required ratio is $$1 : 16.$$

Hence, the correct answer is Option A.