Let in 2s body reaches upto point A and after one more second upto point B.

Total time of ascent for a body is given as 3s



i.e., $\frac{u\sin \theta }{g}=3$

$\therefore$ $u\sin \theta =10\times \mathrm{3 }$$=30$ ......(i)

Horizontal component of velocity remains always constant

$u\cos \theta =v\cos 30°$ .......(ii)

For vertical upward motion between point O and A

$v\sin {30}^o=u\sin \theta -g\times 2$  [Using v = u - gt]

$v\sin {30}^o=30-20$ $\left[As u\sin \theta =30\right]$

$\therefore$ $v=20 \mathrm{m} {\mathrm{s}}^{-1}$

Substituting this value $v$ in equation (ii), we get 

$u\cos \theta =20\cos 30°=10\sqrt{3}$ .....(iii)

From equations (i) and (ii)

$u=20\sqrt{3}$ ${\text{m s}}^{-1}$ and $\theta =60°$