= Solution
Decompose the required runway into successive layers. The first layer has cost $s$ and is needed by all four planes. The next costs $2s$ and is needed by the three medium-or-long planes. The final layer costs $4s$ and is needed only by the longest plane.
For each layer, the first requiring plane in a uniformly random ordering contributes its entire cost. Each of its requiring planes has equal probability of being first. By <linearity of the Shapley value>, divide each layer cost equally among those requiring it. The payments are therefore
$$
\boxed{\phi_{\mathrm{short}}=\frac s4,\qquad\phi_{\mathrm{medium},1}=\phi_{\mathrm{medium},2}=\frac s4+\frac{2s}3=\frac{11s}{12},\qquad\phi_{\mathrm{long}}=\frac s4+\frac{2s}3+4s=\frac{59s}{12}.}
$$
They sum to $7s=c(N)$. The equality of the two medium payments follows from their identical runway requirements; the random-order argument proves the amounts, not just their equality.
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