Rob Gronkowski follows Lavonte David, Shaquil Barrett and Chris Godwin in re-signing with the Tampa Bay Buccaneers for 2021; all four players were key members of the 2020 Super Bowl-winning side; Gronkowski scored two touchdowns in the Super Bowl win over the Kansas City Chiefs
Tuesday 16 March 2021 07:27, UK
The Tampa Bay Buccaneers have re-signed another key member of their Super Bowl-winning side after Rob Gronkowski agreed to a one-year deal worth up to $10m.
Gronkowski came out of retirement to reunite with his former New England Patriots team-mate Tom Brady in Tampa for the 2020 season, which remarkably ended in them both winning yet another Vince Lombardi Trophy.
Gronkowski's new contract will pay him $8m in 2021, with an additional $2m available in incentives.
The tight end's new deal comes after the Buccaneers re-signed linebackers Lavonte David and Shaquil Barrett to two and four-year deals, respectively.
They also placed the franchise tag on receiver Chris Godwin, meaning the Buccaneers' 2021 roster will have a familiar look to it.
Tampa Bay's Super Bowl win was Gronkowski's fourth, with his previous three coming alongside Brady in New England.
The four-time All-Pro recorded modest numbers by his standards last season, with 45 catches for 623 yards and seven touchdowns.
And in the Buccaneers' first three playoff games he totalled just two catches for 43 yards, but he proved his worth in the Super Bowl, making six receptions for 67 yards and two touchdowns to help dethrone the Kansas City Chiefs.
Those scores mean Gronkowski and Brady have now combined for 14 playoff touchdowns across their careers, making them the most productive post-season duo in NFL history.
After being picked in the second round of the 2010 Draft, Gronkowski recorded 521 catches for 7,861 yards and 79 touchdowns for the Patriots before announcing his retirement - albeit temporarily - in 2018.
