Greg Rousseau Height: How Tall Is Greg Rousseau?

Gregory Rousseau was born on April 5, 2000, he is an American football defensive end for the Buffalo Bills of the National Football League (NFL). He is 6 feet 6 inches tall.



He played college football at the University of Miami. As a redshirt freshman there in 2019, he recorded 15.5 sacks and won all-conference honors, and was named ACC Defensive Rookie of the Year. Rousseau was drafted by the Bills in the first round of the 2021 NFL Draft.

Rousseau attended Champagnat Catholic School in Hialeah, Florida. He played defensive end, safety, and wide receiver in high school. He committed to the University of Miami to play college football.



Rousseau played in the first two games of his true freshman year in 2018 before suffering a season-ending ankle injury and being redshirted. He entered his redshirt freshman year in 2019 as a backup but eventually took over as a starter later in the season.

He was named ACC Defensive Rookie of the Year after recording 15.5 sacks that season, second in the nation behind only Chase Young’s 16.5. He opted out of the 2020 season due to concerns regarding the COVID-19 pandemic.

Rousseau was drafted by the Buffalo Bills in the first round (30th overall) in the 2021 NFL Draft. He signed his four-year rookie contract, worth $11.37 million, on June 4, 2021.

During Week 2 of the 2021 season, Rousseau was nominated for NFL Rookie of the Week honors with five tackles, two sacks, and two tackles made behind the line of scrimmage.

Rousseau recorded his first career interception on October 10, 2021, in a 38-20 Week 5 Sunday night win against Patrick Mahomes and the Kansas City Chiefs, earning AFC Defensive Player of the Week.