Rob Gronkowski, Bucs reportedly agree to 1-year deal
Tom Brady will have one of his favorite targets with him again next season as the Tampa Bay Buccaneers defend their Super Bowl championship.
Tight end Rob Gronkowski agreed to a one-year deal to play another season in Tampa Bay, agent Drew Rosenhaus told NFL Network and ESPN on Monday.
The Athletic and NFL Network reported that the contract will pay Gronkowski $8 million and includes the opportunity to earn another $2 million in incentives.
Gronkowski tweeted a gif of Patchy the Pirate from SpongeBob SquarePants saying, “Arrr You Ready Kids?!?”
Gronkowski, 31, came out of retirement before the 2020 season to reunite with his former New England Patriots quarterback, Brady, in Tampa Bay. The Buccaneers’ title last month gave Brady a record seven Super Bowl titles while Gronkowski won his fourth.
A four-time All-Pro, Gronkowski produced mediocre numbers by his standards in the 2020 regular season: 45 catches for 623 yards and seven touchdowns. In the Buccaneers’ first three playoff games, he totaled just two catches for 43 yards, but he was a key contributor in the Super Bowl against the Kansas City Chiefs, making six receptions for 67 yards and two TDs.
The Brady-Gronkowski connection has accounted for 14 career playoff touchdowns, the top figure for any quarterback/pass-catcher duo in league history.
Gronkowski, selected by New England in the second round of the 2010 draft out of Arizona, played for the Patriots from 2010-18, compiling 521 catches for 7,861 yards and 79 touchdowns. He topped the 1,000-yard receiving mark four times.
–Field Level Media