Two seconds after projection a projectile is moving at 30°above horizontal .After one more seconds it will be moving in a horizontal

The horizontal component of velocity is always same. Let velocity at t=2 be v and initially be u.We can say v×cos30°=u× cos(x)X is initial angle of projection.Let this be equation (1)Now from t=2 to t=3Vertical velocity changes from vsin30° to 0.Thus using v=u +atWe get 0=vsin30°-g×1v= 2gPutting this value in equation 12g=u× cos(x)Also time of flight is 3 secThus 2usin(x)/g=3usin(x)=3/2Dividing bith equationsTan(x)=3/4g=3/40From here we have initial angleWe can now calculate any asked question.