Solution:     

Acceleration due to gravity = g = $-10 ms^{-1}$ (for upward motion)
Time to reach maximum height (one sided time) = t =$\frac{6}{2}$= 3 s
Velocity at maximum height = Vf = 0 $ms^{-1}$

(i)Maximum height reached by the ball S = h =?
(ii)The maximum initial velocity of the ball = $V_{i} =?$

 Since,      $V_{f} = V_{i} + g\times t$ 
                 $V_{i} = V_{f} – g\times t$                                
                 $V_{i} = 0 – (-10) \times 3$
                 $V_{i}= 30 ms^{-1}$ 

Now using 3rd equation of motion

                  2aS = $V_{f}^{2} – V_{i}^{2}$ 
                  S = $V_{f}^{2}–\frac{V_{i}^{2}}{2a}$
                  S = (0)2 – $(30)\frac{2}{2}\times(-10)$
                  S = $\frac{-90}{-20}$
                  S = 45 m