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Former New England Patriots tight end Rob Gronkowski is to come out of retirement to reunite with superstar quarterback Tom Brady at the Tampa Bay Buccaneers.
Gronkowski, 30, retired from the NFL in March 2019 – weeks after helping the Patriots to victory over the Los Angeles Rams in Super Bowl 53.
Gronkowski had a year remaining on his contract when he retired so the Pats still held his rights, which they have now traded to Tampa Bay for a fourth-round draft pick, NFL.com reports.
The Patriots will also give up a seventh-round pick as part of the deal for a player considered to be one of the greatest tight ends in NFL history.
Three-time Super Bowl winner Gronkowski has not been idle in retirement and recently performed in the WWE, winning the 24/7 title at Wrestlemania 36 earlier this month.
A five-time Pro Bowl selection, Gronkowski caught 521 passes for 7,861 yards and 79 touchdowns in 115 games from 2010-18.
He added 81 catches for 1,163 yards and 12 touchdowns in 16 postseason contests.
His move to Tampa Bay would be a big story on its own, but reuniting with Brady – with whom he shared so much success across nine seasons in New England – means the media focus on the Bucs next season will be intense.
Brady, 42, joined the Buccaneers in March after 20 years with the New England Patriots. He is regarded as one of the greatest players of all time, having won the Super Bowl six time – more times than any other player – in 2002, 2004, 2005, 2015, 2017 and 2019.
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