Answer
$\theta=0.141^{\circ}$ or $\theta=89.86^{\circ}$
Work Step by Step
Let's apply the equation $S=ut +\frac{1}{2}at^{2}$ in the vertical direction to find the flight time of the projectile. (Let's assume the initial velocity = $V_{0}$, angle with horizontal = $\theta$)
$\uparrow S=ut +\frac{1}{2}at^{2}$ ; Let's plug known values into this equation.
$0=V_{0}sin\theta\times t+\frac{1}{2}(-g)t^{2}$
$t=\frac{2V_{0}sin\theta}{g}-(1)$
Let's apply the equation $S=ut $ in the horizontal direction to find the launch angle $\theta$ of the projectile.
$\rightarrow S=ut $ ; Let's plug known values into this equation.
$R=V_{0}cos\theta\times t=V_{0}cos\theta\times \frac{2V_{0}sin\theta}{g}$
$R=\frac{V_{0}^{2}sin2\theta}{g}=>sin2\theta=\frac{Rg}{V_{0}^{2}}$ ; Let's plug known values into this equation.
$sin2\theta=\frac{91.4\space m\times9.8\space m/s^{2}}{(427\space m/s)^{2}}=4.91\times 10^{-3}$
$2\theta=sin^{-1}(4.91\times10^{-3})=0.281^{\circ}\space or\space 180^{\circ}-0.281^{\circ}=179.719^{\circ} $
$\theta=0.141^{\circ}\space or\space 89.86^{\circ}$
