Sports

LaMelo Ball stays in play for Cleveland regardless of Collin Sexton and Darius Garland's presence

The Cleveland Cavaliers already have two promising young guards on the team in Collin Sexton and Darius Garland. However, it is rumored that, despite the presence of Sexton and Garland on the roster, LaMelo Ball will stay in play for the Cavs whenever the 2020 NBA draft takes place.

During an interview on 92.3 The Fan, Cleveland.com's Chris Fedor said that if the Cleveland Front Office found its way, Ball could land in the Cavs via Lonzo Wire's Jacob Rude:

“The feeling I get is that you have a top group of people and it's three of them. It's James Wiseman, Anthony Edwards, and LaMelo Ball. And I think no matter what the Cavs do, they won't Allow fit and position to determine that decision. You will consider a number of different factors and summarize and weight them. But you will not say, "Because we designed Darius Garland last year and Collin Sexton the year before, LaMelo Ball is out. "They won't do that. You don't feel like you can."

Ball played 12 games in the NBL before sustaining a foot injury. The youngest ball brother averaged 17.0 points, 7.4 rebounds and 6.8 assists per competition.

In the era of the little balls, it would be fascinating to see the Cavs throw out a line-up of Sexton, Garland and LaMelo. The Cavs will have a large selection in the 2020 draft. They had the worst record in the Eastern Conference before the 2019-20 season was suspended due to COVID-19.

The Cavs desperately need young talent in every position. As Fedor says, the organization doesn't have the luxury of being picky right now as it had the second worst record in the NBA before the games were suspended.

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