Tight end Rob Gronkowski reaches another one-year deal with Bucs

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Joey Knight, Tampa Bay Times
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Having successfully kept most of the band together defensively, the Bucs on Monday secured one of their most profound offensive instrumentalists.

Hours after indicating he might be inclined to “dip my toes” into free agency, veteran tight end Rob Gronkowski agreed to another one-year deal with Tampa Bay that unites him with fellow future Hall of Famer Tom Brady for an 11th season. Gronkowski’s agent, Drew Rosenhaus, told ESPN the deal is worth $10 million.

Acquired from the Patriots via trade last April, he worked off the final year of an existing contract in 2020, which paid him $9.25 million according to OverTheCap.com.

Gronkowski, who turns 32 in May, played in all 20 games in 2020 despite coming off a one-year retirement. He finished with 45 catches for 623 yards and seven touchdowns in the regular season, then proved more valuable as a Brady protector than target for much of the playoffs, getting a ball thrown his way only once in wins against Washington and Green Bay.

But in Super Bowl 55, he exploded for six catches for 67 yards and two touchdowns. In the run-up to the Super Bowl, Gronkowski made it clear he wanted to remain with the Bucs in 2021.