NBA announced the last of its individual awards by naming Charlotte Hornets’ LaMelo Ball as the rookie of the year for the 2020-21 season. The point guard enjoyed a prolific first season with the Hornets, who were eliminated in the play-in tournament.
LaMelo Ball led the voting over his fellow rookie finalists Anthony Edwards of Minnesota Timberwolves and Tyrese Haliburton of Sacramento Kings.
“Charlotte Hornets guard LaMelo Ball has been voted the NBA’s Rookie of the Year, sources tell ESPN. Announcement expected soon,”
The youngest of the Ball brothers had to prove his worth by playing in overseas leagues before declaring himself for the NBA Draft. He was picked by the Charlotte Hornets as the third overall draft pick ahead of the season.
Following a rough debut against Cleveland Cavaliers, where he failed to score a single point, Ball slowly adjusted to the level. He went on to become the youngest ever player to record a triple double in the NBA down the season.
His season came to a screeching halt when he suffered a broken wrist in a loss against Los Angeles Clippers. While he was ruled out indefinitely, he returned just in time to help the Hornets make a late surge in the regular season. The franchise finished 10th in the Eastern Conference table to qualify in the play-in tournament. A heavy defeat to Indiana Pacers sealed their exit from the season,
The 19-year-old played 51 games in the campaign, out of which 31 were starts. He averaged 15.7 points, 6.1 assists, and 5.9 rebounds per match.
He has clearly impacted winning, more so than I even expected,” Hornets head coach James Borrego said in early April, via the Charlotte Observer. “When you can impact winning and put up those numbers, you are clearly Rookie of the Year.”