Rob Gronkowski was weighing up whether to split from Tom Brady for the first time in his NFL career before settling on a sweet new deal.
Tampa Bay Buccaneers tight end Rob Gronkowski has re-signed with the Super Bowl champions on a one-year deal worth up to $AUD13 million, reports said.
Gronkowski, a long-time New England Patriots teammate and close friend of Buccaneers quarterback Tom Brady, will receive $10 million for the 2021 campaign with bonuses worth an additional $3 million, NFL.com reported.
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The 31-year-old future Hall-of-Famer came out of retirement last year to reunite with Brady in Tampa Bay, and scored two touchdowns in the Bucs’ 31-9 Super Bowl win over the Kansas City Chiefs last month.
Before signing his extension, Gronkowski had suggested he might be willing to split from Brady for the first time in his NFL career, even though his preferred option was to remain with the Buccaneers.
“With the free agency process, you just never know what may happen,” he told the 10 Questions with Kyle Brandt podcast.
“There’s some other teams interested out there, too. I’ve also never been a part of the free-agency process, so I actually want to dip my toes in, just to see what’s out there.
“I want to be a free agent every year, so if I dip my toes in this year, see what’s out there, see how it works, even if I do go back to the Bucs … I’ll know how free agency works and I’ll know how teams come after you. Then I’ll have that much more of an advantage every single year.”
However, the charismatic footballer decided to stay put with the Bucs and go around with Brady again.
News of Gronkowski’s return to Tampa Bay next season comes after Brady agreed on the weekend to a contract extension that will keep him in Florida for the next two seasons at least, on a deal worth a reported $65 million, per NFL Network’s Tom Pelissero.
Brady’s restructured deal allows Tampa Bay to make payroll savings aimed at ensuring that the core of the team that won the Super Bowl is available to return next season.
As the NFL’s free agency negotiating period opened on Monday, the Buccaneers also moved to lock up star pass-rusher Shaquil Barrett.
Barrett has inked a four-year deal worth up to $93 million, with $46 million in salary guaranteed.
He confirmed the new deal in a post on Instagram. “If it’s meant to be, it will be,” Barrett wrote. “Something special about this whole organisation and am so excited to be a part of it for four more years.”
Barrett was a key member of the Buccaneers’ dominant defence last season, sacking Green Bay quarterback Aaron Rodgers three times in the NFC Championship game before sacking Chiefs star Patrick Mahomes in the Super Bowl.
In other notable free agency deals, the Chiefs strengthened their offensive line after snapping up New England’s Joe Thuney in a five-year deal worth $103 million, according to ESPN.
The 28-year-old Thuney played five seasons for the Patriots after being drafted in 2016, and appeared in three straight Super Bowls in 2017, 2018 and 2019.
The move should give Chiefs quarterback Mahomes better protection after he was run ragged by Tampa Bay in the Super Bowl.