Former Ohio State defensive end Cameron Heyward was named as the NFL's 48th-best player on Monday night when NFL Network continued its countdown of the league's top 100 players, as voted on by players from around the league.
.@steelers versatile DL @CamHeyward comes in at #48 on the list!#NFLTop100 pic.twitter.com/Zsw2yDkiV4
— NFL Network (@nflnetwork) May 29, 2018
Heyward is coming off his seventh season in the NFL, which was also his best season in the league to date, as he recorded 45 total tackles, including 12 sacks, and earned AP All-Pro honors for the first time in his career.
A member of the Buckeyes from 2007-10, Heyward has played his entire NFL career with the Pittsburgh Steelers.
Heyward is one of two Buckeyes-turned-Steelers in this year's rankings, with Ryan Shazier (No. 47) checking in just one spot ahead of him. This year's list includes a total of eight players from Ohio State, the most among all schools, also including Carlos Hyde (No. 97), Malcolm Jenkins (No. 96), Marshon Lattimore (No. 82), Michael Thomas (No. 81) and Ezekiel Elliott (No. 54). The top Buckeye on this year's list has yet to be revealed, with the top 40 players set to be announced in 10-player increments over the next four Monday nights.
