5.5 s
Explanation:
The time it takes for the ball to reach its maximum height can be calculated using
[tex]v_y = v_{0y} - gt \Rightarrow t = \dfrac{v_{0y}}{g}[/tex]
since [tex]v_y = 0[/tex] at the top of its trajectory. Plugging in the numbers,
[tex]t = \dfrac{(54\:\text{m/s})}{(9.8\:\text{m/s}^2)} = 5.5\:\text{s}[/tex]