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Rob Gronkowski returning to Tampa Bay Buccaneers, says agent

NFL, Tampa Bay Buccaneers

The Super Bowl champion Tampa Bay Buccaneers continue to get the band back together. 

The latest to rejoin: Rob Gronkowski. 

The star tight end and the Bucs have agreed to a 1-year deal worth $10 million, his agent, Drew Rosenhaus, told ESPN's Adam Schefter on Monday. 

Gronkowski was one of several high-profile Bucs players -- defensive lineman Ndamukong Suh, wide receiver Antonio Brown, kicker Ryan Succop and running back Leonard Fournette were among the others -- whom the team had hoped to bring back following last month's Super Bowl win.

Earlier Monday, the Bucs and star pass-rusher Shaquil Barrett agreed to a four-year deal worth up to $72 million, Rosenhaus told Schefter. 

The Buccaneers traded a fourth-round draft pick to the New England Patriots last offseason for Gronkowski's rights and a seventh-round pick. Gronkowski, who had retired from the NFL in 2019, started off the season slowly, shaking off some rust after being away from football for a year and finding a role in Bruce Arians' offense, which wasn't known for being tight end friendly.

While offensive coordinator Byron Leftwich and quarterback Tom Brady worked to carve out a role for Gronkowski, he embraced his role as a blocker and as a leader and finished the regular season with 45 receptions for 623 yards and 7 touchdowns.

He peaked at the right time, though, finishing with two touchdowns in the Buccaneers' 31-9 victory in Super Bowl LV as he and Brady broke Joe Montana's and Jerry Rice's postseason record for most touchdowns by a QB-receiving duo.

Gronkowski, 31, spent the first nine seasons of his NFL career with the Patriots, winning three Super Bowls with the team. He was a five-time Pro Bowl selection and four-time first-team All-Pro.

He has 566 receptions for 8,484 yards and 86 touchdowns in 131 regular-season games in his career.

ESPN's Jenna Laine contributed to this report.

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