1. Interpretation of the question

A bullet fired from ground with speed $u$ can be projected at any angle $heta$ with the horizontal.
The set of all points on the horizontal ground that can be reached by the bullet forms a region.

For projectile motion on level ground, the horizontal range is

$R=\frac{u^2\sin 2\theta}{g}.$

Since the gun can fire in all possible directions, the bullet can land at any ground point whose distance from the firing point is at most the maximum range.

So the covered region on the ground is a circle of radius

$R_{\max}=\frac{u^2}{g},$

because maximum range occurs when

$\sin 2\theta = 1 \quad \Rightarrow \quad \theta = 45^\circ.$

2. Area covered on the ground

Hence, maximum area covered by a gun of speed $u$ is

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

Thus,

$A \propto u^4.$

3. Apply to guns A and B

Given:

• Gun A speed: $u_A = 1\ \text{km/s}$
• Gun B speed: $u_B = 2\ \text{km/s}$

Therefore,

$\frac{A_A}{A_B} = \left(\frac{u_A}{u_B}\right)^4 = \left(\frac{1}{2}\right)^4 = \frac{1}{16}.$

So the ratio of maximum areas covered is

$A_A : A_B = 1 : 16.$

4. Option check

• A: $1:16$ ✅
• B: $1:8$ ❌
• C: $1:2$ ❌
• D: $1:4$ ❌

Thus, the correct option is A.