FOXBOROUGH, Mass. - New England Patriots tight end Rob Gronkowski has been awarded first-team All-Pro honors, which earns him $2.5 million and achieves the highest level of incentives possible as part of the revamped contract he signed last offseason.
When the season began, Gronkowski had a chance to earn an additional $5.5 million in incentives based on a combination of postseason honors, playing time and statistical production.
To reach the top tier of incentives, he had to be named first-team All-Pro, play at least 90 percent of the offensive snaps or reach 80 receptions or 1,200 receiving yards.
Gronkowski finished the regular season playing 79.3 percent of the offensive snaps, missing one game because of injury (Oct. 5 at Tampa Bay) and one (Dec. 11 at Miami) for an NFL suspension for his late hit on Buffalo Bills cornerback Tre'Davious White.
He totaled 69 catches for 1,084 yards with eight touchdowns in the regular season.
Gronkowski tweeted about the honor Friday afternoon.
Thank you voters! Hard work and working Smarter is paying off!!!
— Rob Gronkowski (@RobGronkowski) January 5, 2018