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Tight end Rob Gronkowski, who is currently retired from the New England Patriots, has agreed to return to the NFL as a member of the Tampa Bay Buccaneers as part of a trade, according to Gronkowski’s agent.
“Pending the physical, Rob has agreed to play for Tampa this season,” agent Drew Rosenhaus told CNN. “He will honor his current contract at this time.”
Gronkowski has one year left on his current contract, which is worth $10 million.
Tampa Bay declined to comment on the trade. New England has not responded to CNN’s inquiry.
The trade would reunite Gronkowski with quarterback Tom Brady, who signed with the Tampa Bay Buccaneers as a free agent on March 20th after 20 seasons with New England. Gronkowski played nine seasons for the Patriots alongside Brady. The duo were teammates on 3 Super Bowl champion teams in the 2014, 2016, and 2018 seasons, though Gronkowski missed the 2016 Super Bowl due to injury.
Speaking on ESPN’s SportsCenter Tuesday night, Rosenhaus said the discussions began shortly after Brady went to Tampa. Gronkowski and the agent agreed it would be an appealing situation.
“He loves new England. He loves the Patriots organization, Coach (Bill) Belichick, the Krafts, his teammates that are still there. He had an amazing run in New England. It was just time for him to continue his career with Tom Brady in Florida with Tampa,” said Rosenhaus. “It was something cool and exciting and challenging for him. He’s well rested and looking forward to playing with Tom again.”
Gronkowski last played during the 2018 NFL season, where he finished with 682 receiving yards and three touchdowns across 13 regular season games.
His final game as a member of the Patriots was in Super Bowl LIII where the New England Patriots defeated the Los Angeles Rams 13-3.
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