Answer:
The projectile flies 1.93 seconds in the air.
Explanation:
Consider the vertical motion of projectile,
   We have equation of motion s = ut + 0.5 at²
       Initial velocity, u = 9.81 x sin 39 = 6.17 m/s
       Acceleration, a = -9.81 m/s²
       Displacement is zero when the shell reaches back to ground, s = -6.31 m
       Time, t =?
   Substituting,
       s = ut + 0.5 at²
       -6.31 = 6.17  x t + 0.5 x (-9.81) x t²
       4.905 t² - 6.17 t - 6.31 = 0
       [tex]t=\frac{-(-6.17)\pm \sqrt{(-6.17)^2-4\times 4.905\times (-6.31)}}{2\times 4.905}\\\\t=\frac{6.17\pm \sqrt{161.87}}{9.81}\\\\t=\frac{6.17\pm 12.72}{9.81}\\\\t=1.93s\texttt{ or }t=-0.67s[/tex]
Negative time is not possible,
The projectile flies 1.93 seconds in the air.
