Former Cleveland Browns safety has agreed to a one-year contract with the Pittsburgh Steelers, according to Ian Rapoport of NFL Network.
Thornhill was reportedly weighing his options between the Steelers and the San Francisco 49ers, with whom he visited last Thursday. Ultimately, he chose the Steelers, believing it offered him the best opportunity.
Originally drafted by the Kansas City Chiefs in the second round of the 2019 NFL Draft out of Virginia, Thornhill spent four seasons with the Chiefs before becoming a free agent. He then signed a three-year, $21 million contract with the Cleveland Browns. However, he was released by the Browns in February, entering the final year of a deal worth $7 million.
His release is believed to be due to injury concerns, as he missed six games in each of the last two seasons. Additionally, his high cap number raised questions given his injury history.
Thornhill played in 11 games for Cleveland in 2024, recording 49 tackles and three passes broken up. Over his career, he has accumulated eight interceptions, including a pick-six in his rookie season, and has appeared in 87 games, starting 74 of them.
With the Steelers, Thornhill is likely to serve as a backup to Minkah Fitzpatrick at free safety and contribute in nickel packages. DeShon Elliott, who had a strong season at strong safety, is expected to keep his starting role.
This move gives Thornhill the opportunity to face the Browns twice a year, offering him a chance to prove they made a mistake by releasing him. The terms of his contract have not been disclosed.
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