Four-time All-Pro TE Rob Gronkowski staying with Bucs on one-year deal

Four-time All-Pro tight end Rob Gronkowski apparently wasn't all that interested in testing free agency this time around, after all. 

Drew Rosenhaus, Gronkowski's agent, told ESPN and NFL Network late Monday afternoon that his client is signing a one-year, $10 million contract to stay with the Super Bowl champion Tampa Bay Buccaneers. Most expected the 31-year-old would remain teammates with long-time friend and Super Bowl Most Valuable Player Tom Brady, but Gronkowski raised eyes during an appearance on the "10 Questions with Kyle Brandt" podcast when he said he wanted to "dip my toes" in free agency "just to see what’s out there." 

Pro Football Focus had previously predicted Gronkowski would sign a one-year deal worth $10 million. The four-time Super Bowl champion told Brandt he will be signing only one-year contracts for the rest of his playing career. 

NFL Network's Tom Pelissero added that Gronkowski's latest agreement is worth $8 million at the start and includes incentives to get up to the $10 million figure. 

Gronkowski may no longer be the player of old, as he recorded 45 receptions for 623 yards and seven touchdowns during the 2020 regular season coming off his year-long retirement. However, he caught two touchdown passes in Tampa Bay's Super Bowl LV victory over the Kansas City Chiefs. 

Gronk addressed the news via Twitter: