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.
