Since we are neglecting air resistance the energy of the ball is conserved. We set the gravitational potential energy to be U=0 at h=0. Applying conservation of energy at h=0 and h=hmax , we get: U1 + K1=U2 +K2(1), at h=0 the potential energy is U1=0 since we did set it so and at the maximum height the speed is 0, therefore K2=0. So, (1) becomes mv2/2=mghmax (2).
Now applying conservation of energy at h=0 and h=hmax/2: mu2/2+mghmax/2=mv2/2, and using (2) we get, mu2/2 +mv2/4=mv2/2, which simplifies to: u2=v2/2, therefore the speed of the ball at h=hmax /2 is u=v/sqrt(2)=28.28 m/s