The Las Vegas Raiders are adding more depth to their tight end room. On Wednesday, the team signed veteran tight end Austin Hooper to a one-year deal worth a reported $3.5 million, according to NFL Media.
This is the latest move in what has been a busy offseason for the Raiders at the position. First, the team made a notable trade that sent Pro Bowl tight end Darren Waller to the New York Giants in exchange for a third-round pick. Following that deal, the team dipped into free agency to replenish the tight end room by signing ex-Texan O.J. Howard and have now brought Hooper into the fold.
The 28-year-old has bounced around the league in recent years and will now be playing on his third team in as many seasons. He signed a one-year deal with the Tennessee Titans last offseason and appeared in 17 games for the franchise, albeit with just two starts. Hooper saw the least amount of offensive snaps since his rookie season (44%), playing in just 51% of Tennessee snaps. Despite that minimal playing time, Hooper was second on the team in receptions and tied for third in receiving yards.
Before landing in Tennessee, Hooper spent the previous two seasons with the Cleveland Browns after inking a four-year, $44 million deal in free agency back in 2020. That season he was able to help the franchise make it to the postseason, and did catch nine passes for 62 yards and a touchdown during that playoff run. The following year he again was slotted in as the starter in Cleveland but didn't produce to a prolific degree, catching 38 passes for 345 yards and three touchdowns over 16 games. He'd be released that offseason and ultimately latch on with Tennessee.
Hooper initially entered the NFL as a third-round draft choice of the Falcons out of Stanford in 2016. Atlanta was where Hooper enjoyed the most success of his career to this point, being named to consecutive Pro Bowls in 2018 and 2019.