Rob Gronkowski is returning to Tampa Bay.
The star tight end is running it back with the Super Bowl-champion Tampa Bay Buccaneers on a one-year deal worth up to $10 million deal, NFL Network Insider Ian Rapoport reported Monday, via Gronkowski's agent Drew Rosenhaus. The contract is for $8 million with the potential to earn $10 million due to incentives, NFL Network's Tom Pelissero added.
Following a retirement that lasted for just the 2019 season, Gronkowski resumed his NFL career by joining longtime New England Patriots teammate Tom Brady with the Bucs en route to winning the Super Bowl LV.
Since the team's defeat of the Kansas City Chiefs, Gronkowski, 31, has made it known he was planning to return, that he wanted it to be with the Bucs and most recently telling Good Morning Football's Kyle Brandt that he would be a "one-year-deal guy for the rest of my career." All of those things have come to fruition.
Gronkowski, No. 43 in Gregg Rosenthal's Top 101 free agents of 2021, came on late in the 2020 season and produced 45 catches for 623 yards and seven touchdowns. In the Super Bowl, though, the future Hall of Famer showcased the clutch play that's been a hallmark of his career, as he hauled in a pair of touchdowns from Brady.
After 10 NFL seasons, the first nine with the Pats, Gronkowski is returning for his second with the Bucs.
Like just about every Super Bowl winner before them, the Buccaneers followed their triumph by stating their desire to run it back with another title.