Initial velocity of #V=6m/sec# directed at an angle #phi=pi/12# to horizon can be represented as a sum two vectors:

#V_v = V*sin(phi)# directed vertically up and

#V_h = V*cos(phi)# directed horizontally.

It's the vertical component #V_v# that drives the projectile up against the constant gravity deceleration #g=9.8m/sec^2#

Since every second the projectile looses its vertical component by #g#, we can calculate the time until its vertical component becomes zero (the top of trajectory):

#t = V_v/g#

Knowing time to the top and initial vertical velocity, the length is calculated by a formula

#S = V_v*t - (g*t^2)/2 = V_v^2/g - (g*V_v^2)/(2g^2) = V_v^2/(2g)#

Incidentally, this is equal to a time #t=V_v/g# multiplied by an average speed, which under constant deceleration from #V_v# to #0# equals to #V_v/2#.

Substituting the real values for #V_v# and #g#,

#S = [6^2*sin^2(pi/12)]/(2*9.8)=[18*(2sin^2(pi/12))]/19.6=#

#=0.92(1-cos(pi/6)) = 0.92*(1-sqrt(3)/2) =#

# 0.92*0.134 = 0.123 (m)#