AFC Defensive Player of the Week: Can He Do It Again?

Las Vegas Raiders defensive end Maxx Crosby was named the AFC Defensive Player of the Week for his dominant performance in Week Two of the 2024 NFL season. Crosby stood out in the Raiders’ hard-fought 26-23 victory over the Baltimore Ravens, playing a pivotal role in securing the team’s first win of the season. He recorded six total tackles, further proving his consistency and impact on the field. This recognition highlights Crosby’s continued growth as one of the NFL’s elite pass rushers and a leader on the Raiders’ defense.

One of the defining moments of Crosby’s performance was his success against Ravens quarterback Lamar Jackson, one of the most elusive players in the league. Crosby was the only defender to bring Jackson down, which was a key moment in the game. Along with that tackle, Crosby also leads the NFL in tackles for a loss with five through two weeks. His relentless pursuit of the ball carrier kept the Ravens’ offense in check, forcing them to adjust. Late in the fourth quarter, Crosby delivered a crucial sack, helping the Raiders get the ball back and sealing the victory.

With this win, the Raiders improved their record to 1-1, returning from a tough season opener. Crosby’s stellar play, capped by his massive sack in the game’s closing moments, was a driving force behind the team’s success. His leadership on defense, combined with his knack for making game-changing plays, will be essential for the Raiders’ playoff push. As the season progresses, Crosby’s Week Two performance has set a strong tone for what could be an outstanding year for both him and the Raiders.