Pats’ Gronkowski gets mega deal
All-Pro Rob Gronkowski agreed Friday to a $54 million deal with the New England Patriots, the richest contract for a tight end in NFL history.
NEW YORK (AP) — All-Pro Rob Gronkowski agreed Friday to a $54 million deal with the New England Patriots, the richest contract for a tight end in NFL history.
The six-year deal includes $18.17 million guaranteed. It is a stunning move by the team for a player entering just his third NFL season, but the Patriots recognized the game-breaking skills of the record-setting Gronkowski.
“This is a rare deal,” said Drew Rosenhaus, Gronkowski’s agent. “Thanks to Mr. Kraft and Coach Belichick.”
Gronkowski set a league record for the position with 17 touchdown catches in 2011. He also had a record 1,327 yards, and made 90 receptions.
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