Jamal Adams' new deal with the Seattle Seahawks will see him overtake the Denver Broncos' Justin Simmons as the highest-paid safety in the league based on his $17.5m average annual salary.
Friday 20 August 2021 23:23, UK
The Seattle Seahawks and Jamal Adams have agreed a four-year, $70m contract making the All-Pro safety the highest-paid player in his position in the NFL.
NFL Network's Ian Rapoport reports that the deal, which caps negotiations that have been ongoing since the end of last season, includes $38m in guaranteed money.
Adams had been due to make $9.86m in the final year of his rookie contract and now looks set to earn on average $17.5m per year to leapfrog the Denver Broncos' Justin Simmons, who takes home $15.25m on average annually, as the most handsomely-paid safety in the league.
The Seahawks sent the New York Jets two first-round picks to acquire the 2017 draft's No. 6 overall pick last season in a bid to bolster an under-fire defense.
Adams finished the 2020 campaign with 9.5 sacks (a single-season record for defensive backs), one forced fumble, 30 pressures and 83 tackles including 14 for loss having been limited to a career-low 12 games by a groin injury.
The 25-year-old, a first-team All-Pro selection in 2019 and three-time Pro Bowler, did report to training camp this offseason but had been yet to participate in practice or preseason games as of the time of agreeing his new deal amid ongoing talks behind the scenes.
But given his immediate emergence as a leader on Seattle's defense, the organisation never looked likely to enter the season without locking him down.
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