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,
[tex]v^{}-u=-gt[/tex]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,
[tex]\begin{gathered} 0-25\sin (60^{\circ})=-9.8\times t \\ -21.65=-9.8t \\ t=\frac{-21.65}{-9.8} \\ t=2.21\text{ s} \end{gathered}[/tex]Thus, the time of flight of the ball is,
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