Just consider the vertical component $d_y=v_{0,y}t+\frac{1}{2}at^2$ $d_y=(13.9\sin 25\°)t+\frac{1}{2}(-9.8~m/s^2)t^2 \\ \\ d_y=5.87t-4.9t^2$ Set d_y equal to 0 to find the times the ball is on the ground $0=5.87t-4.9t^2 \\ \\ 0=t(5.87-4.9t) \\ \\ t=0 \\ \\ t=1.2~s$ The ball was in the air for 1.2 seconds