= Solution
When an endpoint's horizontal coordinate is fixed but its height is free, the <natural boundary conditions for a free endpoint> require $F_{y'}=0$ there. In this problem,
$$
\boxed{\left.\frac{y'}{y\sqrt{1+y'^2}}\right|_{\text{free endpoint}}=0,
\qquad\text{equivalently }y'=0.}
$$
Every admissible <extremal> is an upper <circular arc> with centre on $y=0$, by part (b). Its tangent is horizontal precisely above its centre. Thus imposing the <natural boundary conditions for a free endpoint> at $x=1$ forces $h=1$. The fixed height at $x=-1$ gives $R^2=4+4c^2$, so
$$
\boxed{y(x)=\sqrt{4(1+c^2)-(x-1)^2},\qquad -1\le x\le1,\quad c>0.}
$$
Its free endpoint height is $y(1)=2\sqrt{1+c^2}$. Positivity on the whole <interval> follows because the radicand is at least $4c^2$.
There can be no solution satisfying the <natural boundary conditions for a free endpoint> at both ends: an upper <circular arc> has only one point with a horizontal tangent, whereas the two endpoints have distinct horizontal coordinates. Therefore
$$
\boxed{\text{There is no admissible extremal with both endpoint heights free.}}
$$
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