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

Given:

Time after projection: 2 seconds

Angle above the horizontal: 30°

Acceleration due to gravity: g = 10 m/s^2

Time to reach maximum height is 3 seconds

$\Rightarrow \frac{u\sin \theta }{g}=3$

First, let's determine the horizontal and vertical components of the projectile's velocity after 2 seconds.

The horizontal component of velocity remains constant throughout the motion because there are no horizontal forces acting on the projectile.

$u\cos \theta =v\cos 30°$

for vertical upward motion between them,

$v\sin 30°=u\sin \theta -g\times 2=30-20$

$v=20m/s$

$u\cos \theta =v\cos 30°=20\cos 30°$

$\frac{u\sin \theta }{u\cos \theta }=\tan \theta =\frac{30}{10\sqrt{3}}=\sqrt{3}$

$\theta =60°$

$u=20\sqrt{3}m/s$

Hence the correct answer is $20\sqrt{3} \mathrm{m}/\mathrm{s}$.