Given data:
* The initial velocity of the ball is 25 m/s.
* The angle of the initial velocity with the horizontal is 60 degree.
* The final vertical velocity of the ball at the maximum height is 0 m/s.
Solution:
By the kinematics equation, the time taken by the projectile to reach the maximum height is,
![v^{}-u=-gt](https://img.qammunity.org/2023/formulas/physics/college/3fj870ymrbfzi0urecwjvp8sa7kxjryn3g.png)
where v is the final vertical velocity at the maximum, u is the initial vertical velocity, g is the acceleration due to gravity, and t is the time taken to reach the maximum height,
![\begin{gathered} 0-25\sin (60^(\circ))=-9.8* t \\ -21.65=-9.8t \\ t=(-21.65)/(-9.8) \\ t=2.21\text{ s} \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/pbtr8d1ifpsc7b1hn6i8ui88yqdbhwxd8r.png)
Thus, the time of flight of the ball is,
![undefined]()