FILE PHOTO: New England Patriots tight end Rob Gronkowski holds up the Vince Lombardi Trophy after his team defeated the Seattle Seahawks in the NFL Super Bowl XLIX football game in Glendale, Arizona, February 1, 2015. REUTERS/Lucy Nicholson
April 21, 2020
(Reuters) – Rob Gronkowski has decided to come out of retirement and will reunite with longtime team mate Tom Brady after the New England Patriots agreed to trade the tight end to the Tampa Bay Buccaneers, according to a report on Tuesday.
Gronkowski, who still has one year remaining on his contract with New England, was a formidable offensive weapon and one of Brady’s most reliable targets at the Patriots where he was part of three Super Bowl-winning teams.
A report on the NFL website sourcing NFL Network Insider Ian Rapoport, said the Patriots agreed to trade Gronkowski and a seventh-round draft pick to the Buccaneers in exchange for a fourth-round pick in the 2020 draft that begins on Thursday.
According to Rapoport, Gronkowski has taken a physical and the player’s agent told him: “Rob has agreed to play for Tampa this season. He will honor his current contract at this time.”
Both the Patriots and Buccaneers have not yet responded to a request for comment from Reuters.
Gronkowski announced his retirement in March 2019, ending one of the NFL’s most dominant and colorful careers after nine seasons. The 30-year-old Gronkowski recently made his debut with World Wrestling Entertainment.
Gronkowski, who spent all nine of his NFL seasons with the Patriots, caught 521 passes for 7,861 yards and 79 touchdowns in 115 games. He added 81 catches for 1,163 yards and 12 touchdowns in 16 postseason contests.
Brady, who won a record six Super Bowl championships with the Patriots, signed with the Buccaneers as a free agent last month and will look to turn around a team that finished 7-9 last year and have not made the playoffs since the 2007 season.
(Reporting by Frank Pingue in Toronto; Editing by Toby Davis)