Rob Gronkowski has decided to end his one-year NFL retirement and return to football.
Gronkowski would be reuniting with his longtime team mate Tom Brady in Tampa Bay Buccaneers. Over the years with the Patriots, Gronkowski and Brady were really close and were both part of the winning team that won three Super Bowls.
While confirming this, Drew Rosenhaus, Gronkowski’s agent said in a statement, “Rob has agreed to play for Tampa this season. He will honor his current contract at this time.”
Gronkowski announced his retirement in March 2019 after is NFL nine seasons with the New England Patriots.
Spending his nine years career with the patriots, Gronkowski caught 521 passes for 7,861 yards and 79 touchdowns in 115 games. In 16 postseason contests he made several more 81 catches for 1,163 yards and 12 touchdowns.
Gronkowski still have one year remaining on his contract with the Patriots with a $9 million salary worth, $750,000 per-game roster bonuses and a workout bonus of $250,000, which all together totals $10 million. His agent however said that Gronkowski would honour his contract with the Patriots.
His decisions to play for Buccaneers in 2020 had led to the need for the two clubs to reach an agreement. As it stands, the Patriots have agreed to trade Gronkowski to Buccaneers in exchange for fourth-round pick. The patriots would also be sending a seventh-rounder to Buccaneers.
