Answer
The minimum initial speed needed is
$$v_0=110.5\text{ m/s}.$$
Work Step by Step
Given the initial speed $v_0$, the maximum range is achieved when the projectile is launched at the $45^\circ$ angle. Conversely, if we have a given range $R$, then the minimum initial speed needed to achieve this range is obtained when the projectile is launched at the $45^\circ$ angle. In this case $R=4086\text{ ft}=1245\text{ m}$ so we have
$$R=\frac{2v_0^2\sin45^\circ\cos45^\circ}{g}=2\frac{v_0^2(\sqrt{2}/2)\cdot(\sqrt{2}/2)}{g}=\frac{v_0^2}{g}$$
which gives
$$v_0^2=gR\Rightarrow v_0=\sqrt{gR}=\sqrt{9.81\text{ m/s}^2\cdot1245\text{ m}}=110.5\text{ m/s}.$$
