Dolphins’ Tua Tagovailoa wins AFC weekly award after historic performance against Ravens

Tua Tagovailoa’s historic passing performance in Sunday’s stunning 42-38 win over the Baltimore Ravens has earned him AFC Offensive Player of the Week honors.

After throwing for 469 yards and six touchdowns to erase a three-touchdown fourth-quarter deficit, the third-year quarterback received the weekly award, announced on Wednesday morning, for the first time in his career.

Tagovailoa’s yardage and touchdown totals Sunday in Baltimore were easily career highs. The six touchdowns tied Hall of Famers Bob Griese and Dan Marino for a franchise record. The 469 yards were fourth-best in Dolphins history, bested three times by Marino.

Four of Tagovailoa’s touchdown passes came in the fourth quarter as Miami, coming back from down 21 points, staged its largest road comeback in franchise history. It was the first time any NFL team came back from down 21 in the fourth quarter in 12 years, according to ESPN Stats & Info.

Tagovailoa’s top two receivers, Tyreek Hill and Jaylen Waddle, collected 11 receptions and two touchdowns each. Hill went for 190 receiving yards and Waddle 171 as Tagovailoa finished 36 of 50.

Tagovailoa is the first Dolphins player to win the AFC’s offensive weekly award since quarterback Ryan Fitzpatrick received the honor in Week 16 of the 2019 season.

Last year, the Dolphins twice had weekly award winners. Cornerback Xavien Howard was the AFC Defensive Player of the Week after his fumble return for a touchdown was a key play in the Week 10 upset of the Ravens. Three weeks later, Michael Palardy took home the conference’s special teams honor for the week for his punting in a victory against the New York Giants.

Tagovailoa’s prolific Sunday has him among NFL passing leaders through two weeks. He’s first in passing yards (739), tied for first in touchdowns (seven), third in yards per attempt (8.9) and completions …read more

Source:: The Denver Post

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