Two second after projection, a projectile is moving at ${30}^{\circ }$ above the horizontal; after one more second it is moving horizontally. Then:

           $u \cos \theta = v \cos 30°$ and     $0 = u \sin \theta – g \times 3,$
$\therefore u \sin \theta = 30 m/s.$
$$\begin{gathered} Also 0 = v \sin 30° – g \times 1, \\ \therefore v = 20 m/s. \end{gathered}$$
Now    $u \cos \theta = 20 \cos 30° = 10 \sqrt{3} m/s.$
∴         u = $\sqrt{{30}^2+{\left(10\sqrt{3}\right)}^2}=20\sqrt{3}$ m/s.
           R = $\frac{2\left(u\cos \theta \right)\left(u\sin \theta \right)}{g}=\frac{2\times 10\sqrt{3}\times 30}{10}$

               = 60$\sqrt{3}$ m.