Before the NBA continues the stretch run of the season in Orlando, a former member of the Golden State Warriors found a new home.
After bouncing around from the Minnesota Timberwolves to the Memphis Grizzlies and the G League, Jordan Bell has agreed to a two-year deal with the Cleveland Cavaliers, according to Shams Charania of The Athletic.
Via @ShamsCharania on Twitter:
Free agent Jordan Bell has agreed to a two-year deal with the Cleveland Cavaliers, sources tell @TheAthleticNBA @Stadium.
— Shams Charania (@ShamsCharania) June 29, 2020
In early February, Bell was traded from the Timberwolves to the Houston Rockets in a blockbuster 12-player deal. The Rockets then moved Bell to the Grizzlies for Bruno Caboclo and a pick swap. The Grizzlies waived Bell after two games. The former Golden State Warriors big man joined the Washington Wizards G League affiliate, the Capital City Go-Go.
Bell was selected No. 38 overall in the 2017 NBA Draft by the Chicago Bulls. The Bulls then traded the Oregon Ducks standout to the Warriors for $3.5 million in cash considerations. In his rookie season with Golden State, Bell averaged 4.6 points on 62.7% shooting from the field, with 3.6 rebounds, 1.8 assists and a block per game. During his rookie campaign, Bell started 13 games with the Warriors.
Since leaving Golden State over the offseason, the Oregon product averaged 3.2 points on 52.2% from the field, with 2.8 rebounds in 8.8 minutes per contest. In his lone G League appearance, Bell tallied 14 points on 7-of-9 shooting from the field with six rebounds, five assists and a block.
With the NBA’s extended transaction window closing on Tuesday, the Cavaliers were able to sign Bell to one of their open roster spots. Bell will compete for a depth role behind Cleveland’s All-Star big man Andre Drummond.
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