Ball is thrown vertically upward and reaches a maximum height of 50 meters. What is the initial speed required

Use the equation V2^2=V1^2 +2a(S1-S2)

When the ball reaches max height, its velocity is 0, the acceleration is -9.8 m/s^2 and the difference in displacement is 50 m, so

0= (V1^2) 2(-9.8 m/s^2)(50m)
0= (V1^2) -980 m^2/s^2
-(V1^2) = -980 m^2/s^2
V1= 31.3 m/s

Use conservation of energy:

1/2 m v^2 = m g h

1/2 v^2 = g h

v = sqrt(2gh)