Seahawks make Adams highest paid safety with four year deal
Seahawks safety Jamal Adams got the contract extension he was looking for on Tuesday, when parties agreed to a four year, $72 million dollar deal
The Seattle Seahawks have made Jamal Adams the highest-paid safety in the NFL, with the team confirming he had signed his extension Tuesday.
Adams inks top deal for safety
ESPN reported Adams has landed a four-year deal with a maximum value of $72million. An average annual value of $17.5m lifted Adams ahead of Justin Simmons ($15.25m).
A resolution between Adams and the Seahawks had long been anticipated, with his previous contract up at the end of the 2021 season.
Seattle gave up two first-round picks in a trade with the New York Jets for the 2019 First Team All-Pro last year, making a long-term deal a necessity.
It's going to be a wonderful journey: Adams
"We officially signed. I'm excited to be here," Adams said in a Twitter post on the Seahawks' page. "It's going to be a wonderful journey, man.
"The next thing on our mind is getting that [championship] – getting right back to it and going to get it."
Adams had a career-high 9.5 sacks in 2020 and ranked second among all safeties in disrupting runs on 15.6 per cent of plays, but the Seahawks were only 16th in defense (364 points allowed).
Brown still waiting for his pay day
While the team will hope Adams can inspire improvement on that side of the ball having agreed terms, there is no resolution in sight for holdout left tackle Duane Brown.
To the frustration of quarterback Russell Wilson, who was sacked 47 times for 301 yards last year, ranking second and third respectively, Brown has not been practicing as he seeks an extension.
Brown has allowed just six of the 146 sacks Wilson has suffered over the past three seasons.