We need to apply the following equation
Vy ^2 = Voy^2 + 2 ay (y - yo)
Where Vy is the final speed of the ball (which will be equal to zero at maximun height)
Voy is the initial speed
a is the acceleration (equal to -9.8 m/s2 ,as it goes against the upward movement)
yo is the initial reference position ( yo = 30m from the ground)
y = yo + [(-Voy^2)/ 2ay] = 30 m + -(20[m/s])^2 / -2(9.8[m/s^2])
y = 30[m] + 20.408 [m] = 50.408 m = 50.4 m
