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.
![\begin{gathered} v_f=v_0+gt \\ t=(v_f-v_0)/(g) \\ t=(0-14.6\cdot(m)/(s))/(-9.8\cdot(m)/(s^2)) \\ t\approx1.49\sec \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/csouhvgndlfv4mhwfvhz0grmst97i4rxuz.png)
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.
![t_{\text{total}}=2\cdot1.49\sec =2.98\sec](https://img.qammunity.org/2023/formulas/physics/college/je3ndyy01g2x0ynyjluyi1i7q2gvzojrra.png)
Therefore, the baseball is 2.98 seconds in the air.