Gronk sticks with Buccaneers for 1 year, $10M

The Tampa Bay Buccaneers will re-sign tight end Rob Gronkowski to a one-year contract for $10 million, agent Drew Rosenhaus told ESPN's Adam Schefter.

The deal matches what Gronkowski played for in 2020 when he came out of retirement to join quarterback Tom Brady in Tampa.

Gronkowski had one of his quietest seasons with the Buccaneers statistically, though the 31-year-old's 623 receiving yards still ranked first among free-agent tight ends. The former New England Patriots star also shined in Super Bowl LV, catching two touchdowns to capture his fourth career championship.

Gronkowski said before the Super Bowl he planned to return to the Buccaneers in 2021, but he acknowledged at the onset of the negotiating period he was looking forward to testing the market for the first time in his career.

With 86 career touchdown catches, Gronkowski ranks third all time among tight ends, trailing only Antonio Gates (116) and Tony Gonzalez (111).

The Buccaneers previously struck agreements to bring back Shaq Barrett and Lavonte David, two of their other prominent free agents.

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