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An alternative construction of such a group: start with the free product of two copies of $A_5$: $G=A_5*A_5$. This group is virtually free, so any torsion-free subgroup is free. Consider the kernel of any group homomorphism from $G$ to $A_5$ that is an isomorphism when restricted to each of the two copies of $A_5$. This kernel is free of rank 59, and since $G$ does contain copies of $A_5$ that map isomorphically to the quotient $A_5$, $G$ is isomorphic to $F_{59}\rtimes A_5$.