At the maximum height of the ball, the velocity will be zero. This will also take half the total time, 3.125 s.
Now use the equation...
Δd = vf*t - 1/2at^2
By setting the final time at the moment of the apex of the height, vf becomes 0.
So the equation will become...
Δd = - 1/2at^2
The equation now resembles the change in height from the ground to the maximum height
Δd = - 1/2at^2 = - 1/2 (-9.8 m/s^2) (3.125 s)^2 = 47.8 m