Two second after projection, a projectile is moving at ${30}^0$ above the horizontal; after one more second it is moving horizontally. Then: $\left(g=10m/s^2\right)$

$u\cos \theta =v\cos {30}^0$

and  $0=u\sin \theta -g\times 3$

$\therefore u\sin \theta =30m/s$

also   $0=v\sin {30}^0-g\times 1,v=20m/s$

Now   $u\cos \theta =20\cos {30}^0=10\sqrt{3}m/s$

$\therefore u=\sqrt{{30}^2+{\left(10\sqrt{3}\right)}^2}=20\sqrt{3}m/s$

$R=\frac{2\left(u\cos \theta \right)\left(u\sin \theta \right)}{g}=\frac{2\times 10\sqrt{3}\times 30}{10}=60\sqrt{3}m$