Seahawks safety Jamal Adams. (Abbie Parr/Getty Images)
The Seattle Seahawks are prepared to offer Jamal Adams a monster contract that will make him the highest-paid safety in the NFL.
The Seahawks want to keep Jamal Adams around, so they’re preparing to make him an offer he can’t refuse.
Seattle is ready to make Adams the highest-paid safety in the league, according to Mike Garafolo of NFL Network.
That would mean topping the four-year, $61 million contract the Broncos recently gave to Justin Simmons.
Simmons will make $15.25 million per year over his four-year deal. Adams will expect to make more than that. The only hiccup is whether or not he makes a stink about his importance being comparable to a linebacker, a position that commands bigger contracts at the top end.
The Seahawks want to make Jamal Adams the highest-paid safety
Ian Rapoport added to Garafolo’s reporting, saying that the Seahawks don’t want to pay Adams more than linebacker Bobby Wagner, who makes $18 million per year.
Something between $15.5 and $18 million will have to be enough to make Adams happy.
Seattle spent quite a bit to bring Adams to the west coast. They sent two first-round draft picks to the Jets to make a trade happen in 2020. They did that knowing full well the safety would want a sizable new contract.
The details still need to get worked out, but it looks fairly clear the Seahawks are going to do what it takes to make sure their draft picks weren’t spent on a rental.
Adams is entering his fifth season in the league with three Pro-Bowl selections to his name. The Jets picked him sixth overall in the 2017 NFL Draft. He had 9.5 sacks and 83 tackles in 12 games during his debut season for Seattle in 2020.
Seahawks fans should be able to look forward to more years of watching him dominate on defense.