Las Vegas Raiders sign TE Austin Hooper to one-year deal
Announced on Wednesday, the Las Vegas Raiders sign TE Austin Hooper to a one-year contract worth $3.5 million. The unrestricted free agent and two-time Pro Bowler will be heading into his eighth NFL season.
Originally selected by the Atlanta Falcons as the 81st overall pick in the third round of the 2016 NFL Draft, Hooper has played in 105 career games while starting in 59 and has recorded 399 receptions for 3,468 yards and two touchdowns. The 6-foot-4, 254 pound tight end has also played for the Cleveland Browns from 2020 to 2022 and the Tennessee Titans last season.
With the Titans, Hooper played in all 17 games of the regular season while starting in two of them. Within that timeframe he collected 41 receptions for 444 yards and two touchdowns. Since 2016, his 339 receptions rank fourth among tight ends in the NFL, and in six of his seven NFL seasons, he has scored at least three touchdowns. Definitely a nice addition as the Raiders sign TE Austin Hooper and lock in a vet with some playoff experience during his 2016 and 2020 seasons.
The San Mateo, California native, played in 27 games over the course of his two seasons at Stanford University and hauled in 74 catches for 937 yards and eight touchdowns. Hooper was a finalist for the Mackey Award and named to the All-Pac-12 first team as a sophomore.
As the Las Vegas Raiders sign TE Austin Hooper, they add in more depth at the tight end position along with O.J. Howard and Jesper Horsted following Darren Waller’s departure.
