Atlanta Braves select contract of Austin Riley
Mississippi Braves alum and Mississippian set to make MLB debut
PEARL, MS. (WLBT) - The Atlanta Braves selected the contract of INF/OF Austin Riley on Wednesday afternoon. The Hernando, Mississippi native will become the 143rd M-Braves alum and third this season to debut in major league baseball since 2005. Riley joins C Alex Jackson (4/7/19) and RHP Jacob Webb (4/16/19) as MLB debuts this season.
Over 75 games for the M-Braves in 2017 and 2018, Riley hit .321 with 19 doubles, four triples, 14 home runs, 47 RBI and a .391 OBP. This season, Riley was currently leading the International League for Triple-A Gwinnett, with 15 home runs and 39 RBI. His grand slam on Tuesday was his 13th home run in his last 18 games and his 10th in May, tying the Gwinnett record for most homers in one month.
Riley was named the Hank Aaron Award winner for Braves Minor League Player of the Year in 2018 after splitting the season between Triple-A Gwinnett and Mississippi. He hit a combined .294 (120-for-408) with 19 home runs and a .882 OPS in 108 games. He began the year with the M-Braves and hit .333 (33-for-99) with a 1.071 OPS in 27 games before a promotion to Gwinnett on May 7.
Riley was first promoted to Mississippi on July 13, 2017 and slashed .315/.389/.511 over 48 games. Riley ranked second in the Southern League in hits (56), tied for fifth in homers (8), and ranked eighth in RBI (27) from July 13 through the end of the 2017 season.
The Atlanta Braves selected Riley in the Competitive Balance A round (41st overall) of the 2015 draft out of DeSoto Central High School. Riley was an outstanding pitcher and hitter in high school, committing to play college baseball at Mississippi State before signing with the Braves. Riley hit .423 during his senior season with 11 home runs and 31 RBI, leading the Jaguars to the 6A state title. On the mound, Riley was 7-2 with a 2.70 ERA, striking out 82 and walking just 19.
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