Tampa Bay Buccaneers: Breakout Candidate – OJ Howard
The Tampa Bay Buccaneers have loads of great players on both the offense and defense. However, one player, in particular, I have slated for a huge breakout season. That player being none other than their first round selection OJ Howard. Howard has the potential to be the Rob Gronkowski of the Buccaneers. It’s easy to see the comparisons, both stand at 6’6”, both are possession and big play tight ends, and they have similar 40 times. The only thing that could put Howard over Gronk going forward is his health. Gronkowski has obviously had his fair share of injuries throughout his career and even coming into the NFL it was one of the bigger concerns. That being said Howard could possibly be one of the best tight ends in the league if he can stay healthy and play like he did at Alabama.
It’s kind of odd to do a breakout piece on a rookie in the NFL especially when you don’t entirely know what they’re capable of. However, OJ Howard looks to be the real deal and could look even better in that offense. An offense in which could be the best. Howard may see a ton of targets this year due to all of the other prolific receivers on this team. You add him to a receiving corps with Mike Evans, DeSean Jackson, fellow rookie Chris Godwin, other tight ends Cameron Brate, and Alan Cross. Howard may kind of get lost in the scuffle at times but he’ll only be missed a couple of times until teams realize they can’t just put single coverage on him. Most linebackers will have a tough time covering him and a lot of defensive backs may have a tough time bringing him down.
Overall I could see OJ Howard finishing top three in the rookie of the year race behind only Christian McCaffrey and Leonard Fournette. He could easily have that big of a season and then some. If Gronkowski continues to struggle with his injuries we may see a new top dog at the tight end position and he may reside in Tampa Bay.