C.J. Stroud torched Baltimore and took home more hardware!
HOUSTON — Houston Texans quarterback C.J. Stroud has been named the AFC Offensive Player of the Week following a stellar performance in the Texans’ 44–10 blowout victory over the Baltimore Ravens, the NFL announced Wednesday. The honor marks the third time in Stroud’s young career that he’s earned the weekly award — and perhaps his most impressive showing yet.
Stroud was nearly flawless in Baltimore, completing 23 of 27 passes (85.2%) for 244 yards and four touchdowns, producing a 143.9 passer rating — the second-highest mark in the NFL for Week 5. His four touchdown passes tied for the most in the league, and his completion percentage was both a personal best and second overall among all quarterbacks that week. The third-year signal caller also showcased his athleticism, breaking free for a 30-yard run — the longest of his career — before capping that same drive with a 10-yard scoring strike to wide receiver Nico Collins.
The Texans’ 44 points tied for the third-highest total by any team this season, and it was the most ever scored by Houston on the road, securing the franchise’s first victory in Baltimore. Stroud’s poise and efficiency were key in orchestrating that milestone win.
Through five games this season, Stroud has completed 70.8% of his passes (102-of-144) for 1,076 yards and eight touchdowns — ranking fourth in the AFC in both categories. Over his last two outings, he’s been on an even hotter streak, completing 45 of 55 passes (81.8%) for 477 yards, six touchdowns, and a blistering 139.2 passer rating — the best in the league during that stretch. Stroud’s command, accuracy, and leadership continue to cement him as one of the NFL’s brightest young stars.
