What makes the projectile to land is the gravity. This is a purely vertical component of the motion.
So we are interested in what the projectile does on the vertical axis.
The velocity vector forms an angle of pi/3 with the ground, so it has the vertical component given by
v_v=9*sin(pi/3)=9sqrt(3)/2\approx7.79 m/s.
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The equation of motion for an object with free fall is
s=s_0+v_0t+1/2g t^2
where s_0 is the initial position, for us it is zero, v_0 is the initial velocity, for us it is v_v, g is the acceleration of gravity -9.81 m/s^2 and t is the time. I use the negative sign for the acceleration because I consider the velocity positive going upward, so the acceleration is negative because it goes downward.
Substituting these values we have
s=7.79t-1/2 9.81t^2
We are interested in the position s=0 because we want to know when the body arrive to the ground (that is the origin of our system of coordinates), so we have
0=7.79t--4.9051t^2.
One solution is t=0 and we expect that at zero time the projectile is still on the ground, then we are interested in the solution with t\ne0. Because of this, we can divide both sides for t and obtain
0=7.79-4.9051t
and the time is then
t=-7.79/-4.9051\approx1.59 s.
