You can use the energy conservation law to do it.

Ep + Ek = constant

0 + (1/2)m v^2 = mgh + 0, for maximum potential energy

Solve for h,

h = v^2/(2g) = 20^2/(2*9.8) = 20.4 m

Answer: At h = 20.4 m the gravitaional potential energy of the ball reaches maximum.

The highest the ball can go is the value that it has the largest gravitational potential energy. You may either calculate it outright , Vf2-Vo2 = 2gh or use the change in KE = the change in PE (1/2 m(Vf2-Vo2) = mgh ).

Either way the answer is the same, 20 m ( if you say g = 10 m/s2 0 or 20.4 m ( if g = 9.8 m/s2 ).

Since there are no retarding forces 

Mechanical energy is conserved.

KE1+PE1=KE2+PE2

 KE2=0 PE1=0 so we get KE1=PE2  1/2mv1^2=mgh2 so we get h2=0.5 v1^2/g=20.4 metersabove ground where potential energy is max.