Eighteen months after departing Wollongong, LaMelo Ball has made his mark on the NBA, after being named the league's rookie of the year on Thursday morning (AEDT).
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The former Illawarra Hawks guard was presented with the honour after a breakout inaugural campaign with the Hornets. He is the third Charlotte player to win the award, following in the footsteps of Larry Johnson in 1991-92 and Emeka Okafor in 2004-05.
Ball represented the Hornets on 51 occasions this season (with 31 starts), averaging 15.7 points, 5.9 rebounds, 6.1 assists and 1.6 steals per game. He receives the honour ahead of Minnesota rookie Anthony Edwards and Sacramento's Tyrese Haliburton.
On NBA on TNT, Ball was asked if his "unconventional" route to the NBA had helped or hindered his development. His answer?
"I'm definitely going to say help me. I think it just fits me, it's who I am. For real," he said.
"Way before the NBA stuff, I thought rookie (of the year) was one of the accomplishments that I could get."
Read more: Hawks new guard grab major NBL gongs
Ball arrived in Wollongong after signing as an NBA Next Star with the Hawks in June 2019. He played 12 games in his solo NBL season, averaging 17 points and 6.88 assists per game, before departing the Illawarra in January 2020 after weeks of speculation about his playing future.
Ball's highlights included becoming the youngest player in NBL history to achieve triple-double (32 points, 13 assists and 11 rebounds) in the Hawks' overtime win over Cairns at the WEC.
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