NFL Draft: Browns select DE Jordan Elliott with 88th overall pick

NFL Draft

COLUMBIA, MO – SEPTEMBER 21: Jordan Elliott #1 of the Missouri Tigers sacks Ryan Hilinski #3 of the South Carolina Gamecocks in the fourth quarter at Faurot Field/Memorial Stadium on September 21, 2019 in Columbia, Missouri. Missouri won, 34-14. (Photo by David Eulitt/Getty Images)

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CLEVELAND (WJW) — After trading down a second time on day two of the NFL Draft, the Cleveland Browns finally made a pick in the third round selecting Missouri DE Jordan Elliott.  He was the 88th overall pick after the team traded down 14 spots with the New Orleans Saints.

Elliott redshirted the 2017 season, earning the team’s defensive Scout Player of the Year Award.  In his Sophomore season, he recorded 24 tackles, 8 for a loss, and three sacks in 13 games and he was not starting.  In his Junior season, Elliott received second-team Associated Press All-American honors after leading his team with 10 tackles for a loss, including three sacks, among 44 total stops in 12 starts.

The Browns traded down 14 spots before selecting Elliott.  Cleveland picked up a third-round pick next year as a result of moving back and the team traded away their 7th round pick in this year’s draft.  The Browns have another pick coming up in the third round at 97 overall.

**Continuing coverage of the NFL Draft**

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