Two seconds after projection, a projectile is traveling in a...

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., $\dfrac{usin \theta}{g} =3$

$∴usin \theta = 10 \times 3= 30$......(i)

Horizontal component of velocity remains always constant

$ucos \theta = vcos 30 °$.......(ii)

For vertical upward motion between point O and A

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

$vsin {30}^{o} = 30 - 20 \left[Asusin \theta =30\right]$

$∴v = 20ms^{- 1}$

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

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

From equations (i) and (ii)

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