Rob Gronkowski is sticking with the Tampa Bay Buccaneers in hopes of getting at least one more crack at another Super Bowl title.
The tight end signed a one-year deal with the Buccaneers on Monday at the start of the NFL’s free agency period, his agent Drew Rosenhaus told ESPN. The deal is reportedly worth $10 million.
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Tampa Bay acquired the rights to Gronkowski in an April 2020 trade with the New England Patriots – about a month after signing Tom Brady. He sat out the entire 2019 season to recover from some injuries and the Buccaneers only had to give up a 2020 fourth-round pick for him.
Gronkowski joined a team that was already loaded offensively, especially at tight end with Cameron Brate and O.J. Howard. As he started to get his feet under him in 2020, he recorded 45 catches for 623 yards and seven touchdowns. He played in all 16 games for the first time since the 2011 season.
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The 10-year pro, as he has in the past, made a significant inpact at this year’s Super Bowl.
Brady found the 31-year-old vet six times for 67 yards and two touchdowns in the 31-9 blowout of the Kansas City Chiefs. The two set a record for most touchdowns by a quarterback-receiver duo in the postseason, breaking Joe Montana and Jerry Rice’s record.
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Tampa Bay has made a significant effort to retain its players that helped them get a Super Bowl LV. The team re-signed Shaq Barrett earlier in the day.
