This diagram shows the dynamics of the situation. $V_j$ is Jack's speed, $V_k$ is Ken's speed. $g$ is their initial goal line separation.
Defining $D$ as the distance between Joe and Ken, we will try to find the minimum of $D^2$ and hence the minimum of $D$
$D^2 = (g-V_j t cos\theta)^2 + t^2(V_k - V_j sin\theta)^2 $
$D^2 = g^2 - 2 g V_j t cos\theta +(V_j^2 + V_k^2)t^2 - 2 V_j V_k t^2 sin\theta$ ..............(1)
The minimum occurs when:
$\frac{\partial{D^2}}{\partial \theta}$ $= 2gt V_j sin\theta - 2 V_j V_k t^2cos\theta =0$ .................(2)
$\frac{ \partial{D^2}}{\partial{t}}$ $= -2g V_j cos\theta +2t(V_j^2 + V_k^2) - 4t V_j V_k sin\theta =0$ ...............(3)
From (2),
$t=$ $\frac{g}{V_k}$ $tan\theta$ ..............................(4)
Substituting (4) into (3):
$ -2g V_j cos\theta +2\frac{g}{V_k} (V_j^2 + V_k^2)\frac{sin\theta}{cos\theta} - 4g V_j\frac{sin^2\theta}{cos\theta} =0$ .....................(5)
Defining $s=sin\theta$ and multiplying (5) through by $\frac{- cos\theta}{2gV_j}$ :
$ (1-s^2)-(\frac{V_j}{V_k}+\frac{V_k}{V_j})s+2s^2= (s-\frac{V_j}{V_k})(s-\frac{V_k}{V_j}) = 0$
If $V_j < V_k$, this quadratic has one solution satisfying $s\lt1$, namely $s=\frac{V_j}{V_k}$
so
$sin\theta=$$\frac{V_j}{V_k}$ and $t=$ $\frac{g}{\sqrt{1-V_j^2/V_k^2}}$.
Substituting these into (1) gives:
$D^2 = $$\frac{g^2}{V_k^2}$$(V_k^2-V_j^2)$
so
$D = g\sqrt{1 - V_j^2/V_k^2}$ ............................... (6)
The notable aspect of this analysis is that to minimise the distance between them when $V_j \lt V_k$, Joe runs at an angle $\theta$ where $sin\theta=\frac{V_j}{V_k}$. When $V_j \gt V_k$, Joe would run at an angle $\theta$ where $sin\theta=\frac{V_k}{V_j}$ in order to intercept Ken.
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