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.
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.
