Rob Gronkowski isn't going anywhere.
The Tampa Bay Buccaneers have agreed to a one-year contract with Gronkowski, according to NFL Media's Ian Rapoport. The deal is worth up to $10 million.
Gronkowski, 31, is coming off his first year in Tampa Bay as he and quarterback Tom Brady helped lead the Buccaneers to a win in Super Bowl LV. The Bucs already have brought back key pieces from that run including receiver Chris Godwin (franchise tag), defenders Shaq Barrett and Lavonte David along with the veteran tight end.
Gronkowski tweeted a hilarious meme to confirm the deal Monday.
The 6-foot-6, 268-pound specimen started all 16 games last season and had another productive campaign. He finished with 45 receptions on 77 targets with 623 yards and seven touchdowns during the regular season.
He dialed back the clock during Super Bowl LV against the Kansas City Chiefs as he finished with six receptions on seven targets for 67 yards and a pair of touchdowns.
Gronkowski, obviously, was a longtime star for the New England Patriots. He spent nine seasons in New England before the Patriots traded his rights to the Buccaneers during the 2020 offseason.
