The former San Francisco 49ers, Oakland Raiders, and Dallas Cowboys standout is set to join the Seahawks as a free-agent. According to Todd Archer, he’s signing a one-year deal. It is his second year in a row playing on a one-year contract, after making $2 million with the Cowboys in 2020.
Last season was his first playing in the NFL since 2015. That year, he was released by the 49ers after being involved in a hot-and-run, during which he was allegedly under the influence. It was his third DUI-related incident since 2012. He signed with the Raiders, and played half the season but was eventually suspended for a full year by the league.
Over the next few years, he was involved in further incidents, including a 2018 allegation of domestic violence. It wasn’t until 2020 that he received reinstatement from the league, and played in all 16 games for the Cowboys. He finished the year with 48 total tackles and five sacks.
Aldon Smith has agreed to a one-year deal with the Seattle Seahawks, according to a source. Smith had his best game of the 2020 season against Seattle, recording three sacks in Week 3. The Cowboys nearly traded him to the Seahawks at the trade deadline but decided to keep him.
— Todd Archer (@toddarcher) April 15, 2021
So far, contract details for Smith have not been disclosed. Even with a productive 2020, it seems unlikely that it is a hugely lucrative deal given Smith’s significant baggage.
As Archer notes, Seattle may be a bit colored by Aldon Smith’s impressive outing against the Seahawks in Week 3. He sacked Russell Wilson three times, the majority of his total for the full season. He had his first sack of the season in Week 1 against the Los Angeles Rams, and only had one more after the Seahawks game: in Week 8 against the Philadelphia Eagles.
During the year, the Seattle Seahawks also unsuccessfully attempted to trade for Smith. Now, they have their guy.