Two seconds after projection, a projectile is moving at $30°$ above the horizontal. After one more second it is moving horizontally. If $\text{g}=10{\text{ms}}^{-2}$ , the velocity of projection is

$\tan \alpha =\frac{\text{u}\sin \theta -\text{gt}}{\text{u}\cos \theta }$

$\Rightarrow \text{Tan}30°=\text{Tan}\theta -\frac{2\text{g}}{\text{u}\cos \theta }$

$\frac{1}{\sqrt{3}}=\tan \theta -\frac{2g}{u\cos \theta }\to \left(1\right)$

After one second it becomes horizontal $\therefore \alpha =0$

$\text{O}=\tan \theta -\frac{3g}{u\cos \theta }\to \left(2\right)$

$\tan \theta =\frac{3g}{u\cos \theta }$

$\frac{3g}{u\cos \theta }-\frac{2g}{u\cos \theta }=\frac{1}{\sqrt{3}}$

$\frac{g}{u\cos \theta }=\frac{1}{\sqrt{3}}\Rightarrow u\cos \theta =10\sqrt{3}$

$u=\frac{10\sqrt{3}}{\cos \theta }\to \left(3\right)$

Sub in (2) we get

$\tan \theta =\frac{3g}{\frac{10\sqrt{3}}{\cos \theta }\times \cos \theta }=\frac{30}{10\sqrt{3}}=\sqrt{3}$

$\tan \theta =\sqrt{3}\Rightarrow \theta =60°$

$u=\frac{10\sqrt{3}}{\cos 60}=20\sqrt{3}{\text{ms}}^{-1}$