Former Ohio State running back Ezekiel Elliott is staying in the National Football League, signing a one-year deal with the New England Patriots on Monday that can be worth up to $6 million with incentives. The contract, first reported by NFL Network’s Ian Rapoport and Tom Pelissero, includes a base salary of $3 million along with a $1 million signing bonus. Elliott is expected to wear No. 15 for New England, the same number he wore at Ohio State for three seasons.
Elliott comes to the Patriots after spending his first seven seasons with the Dallas Cowboys, where he racked up 8,262 total rushing yards and 68 touchdowns. The former Ohio State tailback’s most lucrative years came early in his NFL career, where he was named AP First-Team All-Pro as a rookie in 2016 and earned Pro Bowl honors in three of his first four seasons (2016, 2018, 2019).
The No. 4 overall selection in the 2016 NFL Draft, Elliott put forth an historic three-year career at Ohio State. Serving as head coach Urban Meyer’s lead running back from 2013-2015, the St. Louis, Mo. native recorded 3,961 career rushing yards, a number that currently ranks third all-time among all Ohio State rushers. His 43 total rushing touchdowns also ranks tied for fourth in Ohio State history with former teammate J.T. Barrett. Elliott is most widely recognized for his role in the Buckeyes’ 2014 national championship run, where he rushed for a combined 476 yards and six touchdowns in the team’s College Football Playoff Semifinal victory against Alabama and CFP National Championship win against Oregon.
Now, Elliott will look to reignite his career playing for head coach Bill Belichick in New England, where he will compete for touches with Patriots’ running backs Rhamondre Stevenson, Ty Montgomery II and Pierre Strong Jr., among others.
