Givens.
• Initial speed = 14.6 m/s.
,• FInal speed = 0 m/s (at highest point).
,• Gravity = 9.8 m/s^2.
First, find the time needed to reach the highest point.
[tex]\begin{gathered} v_f=v_0+gt \\ t=\frac{v_f-v_0}{g} \\ t=\frac{0-14.6\cdot\frac{m}{s}}{-9.8\cdot\frac{m}{s^2}} \\ t\approx1.49\sec \end{gathered}[/tex]It takes 1.49 seconds to reach the highest point.
The time that the baseball takes to reach the ground is double t because the trajectory is symmetrical, that is, it takes the same time to go from ground level to highest point than from highest point to ground level.
[tex]t_{\text{total}}=2\cdot1.49\sec =2.98\sec [/tex]Therefore, the baseball is 2.98 seconds in the air.