Hornets Guard LaMelo Ball Wins NBA Rookie Of The Year Award
Jun 16, 2021, 5:41 PM | Updated: Dec 7, 2022, 4:02 pm
LaMelo Ball #2 of the Charlotte Hornets brings the ball up court during the third quarter of their game against the Denver Nuggets at Spectrum Center on May 11, 2021 in Charlotte, North Carolina. (Photo by Jared C. Tilton/Getty Images)
(Photo by Jared C. Tilton/Getty Images)
SALT LAKE CITY, Utah – The NBA announced that Charlotte Hornets guard LaMelo Ball was voted as the league’s Rookie of the Year for the 2020-21 season.
Ball received the honor on Wednesday, June 16.
The Hornets star received more votes than fellow rookies Anthony Edwards (Minnesota Timberwolves), Tyrese Haliburton (Sacramento Kings), and Saddiq Bey (Detroit Pistons).
The 2020-21 #KiaROY is… LaMelo Ball! #NBAAwards #ThatsGame pic.twitter.com/TZc3xxcAGm
— NBA (@NBA) June 16, 2021
After playing professionally in Australia, Ball was selected by the Hornets with the No. 3 overall pick in the 2020 NBA Draft.
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During his rookie campaign, Ball averaged 15.7 points per game on 43.6 percent shooting.
Ball scored a career-high 34 points during a loss to the Utah Jazz on February 5.
In addition to his scoring, the guard recorded 5.9 assists, 6.1 assists, and 1.6 steals per contest.
It's officially official. @MELOD1P IS YOUR 2020-21 @NBA ROOKIE OF THE YEAR! 🕺 🛸 💕
🔗: https://t.co/B1enxDPYEu pic.twitter.com/fnpe3tk7Pt
— Charlotte Hornets (@hornets) June 16, 2021
Ball helped the Hornets to a 33-39 record during the regular season. That mark ranked 10th in the Eastern Conference Standings.
Charlotte suffered a loss to the Indiana Pacers in the NBA Play-In and failed to reach the postseason.
