Its knots!

Here, for taking the motions of the two ships along the x and y axes is going to make us find the integrals,

$\displaystyle\int\limits_{0}^{t} v_0 \cos\theta dt$

$\displaystyle\int\limits_{0}^{t}( v_0 - v_0 \sin\theta )\mathrm{d}t$

But finding them would be tough! Hence we take an easier path..

Trajectroy

Let the vertical axis be $y- axis$

Along the $y- axis$ , the initial distance between the ships is $0$ and final distance is $x$

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Velocity of B along $y -axis = v_0 \sin\theta$ And hence, velocity of B w.r.t A along $y-axis$ ,

$v_{(B-A)y-axis} = v_{(B)y-axis} - v_{(A)y-axis} = v_0\sin\theta - v_0$

$\displaystyle x = \int\limits_0^x \mathrm{d}y = \int\limits_0^t v_{(B-A)y-axis}\mathrm{d}t = \int\limits_0^t (v_0\sin\theta - v_0)\mathrm{d}t\rightarrow\boxed{1}$

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Velocity of A along trajectory = $v_0 \sin\theta$

And hence, velocity of B w.r.t A along trajectory,

$v_{(B-A)trajectory} = v_{(B)trajectory} - v_{(A)trajectory} = v_0 - v_0\sin\theta$

$\displaystyle \int\limits_{16}^x \mathrm{d}s = \int\limits_0^t v_{(B-A)trajectory}\mathrm{d}t = \int\limits_0^t ( v_0 - v_0\sin\theta)\mathrm{d}t \rightarrow\boxed{2}$

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Now, carefully notice that

$v_{(B-A)trajectory} = -v_{(B-A)y-axis}$

i.e.

$\displaystyle\int\limits_0^t ( v_0 - v_0\sin\theta)\mathrm{d}t= -\int\limits_0^t (v_0\sin\theta - v_0)\mathrm{d}t$

Substituting $\displaystyle\boxed{1}$ and $\boxed{2}$ ,

$\displaystyle\int\limits_{16}^x \mathrm{d}s = -\int\limits_0^x \mathrm{d}y$

Or,

$x - 16 = -x$

Notice one thing,

If $d$ is the initial distance,

$\boxed{x = \cfrac{d}{2}}$

$\Rightarrow \boxed{x = 8}$