Tampa Bay Buccaneers

O.J. Howard Nominated for Pepsi NFL Rookie of the Week

The rookie tight end was nominated on Tuesday for the Pepsi NFL Rookie of the Week award.

Pictures of TE O.J. Howard from Sunday's matchup against the Bills.

The NFL announced the honors following a breakout game against the Buffalo Bills this past Sunday that saw rookie tight end O.J. Howard score not only one, but two touchdowns on the day.

In a Week 7 matchup against one of the stingiest defenses in the NFL, quarterback Jameis Winston connected with Howard on a seven-yard touchdown grab in the third quarter for Tampa Bay's first touchdown on the day. The tight end wasn't done yet, as he nabbed a 33-yarder down the sideline for his second score. He finished the day with six catches for 98 yards.

Howard is averaging 18.6 yards a catch this season, with 205 yards on 11 receptions. He was the Buccaneers' first-round pick (19th overall) in the 2017 NFL Draft out of Alabama.

Click here to vote.

This article has been reproduced in a new format and may be missing content or contain faulty links. Please use the Contact Us link in our site footer to report an issue.
Advertising