Frances Tiafoe beats India’s top singles player in Australian Open first round; Kudla advances, too
By Baltimore Sun staff
Jan 14, 2019 | 2:40 AM
Frances Tiafoe of Riverdale in Prince George’s County defeated India’s top-ranked singles player, Prajnesh Gunneswaran, in straight sets in the opening round of the Australian Open on Sunday night, 7-6 (9), 6-3, 6-3, at Melbourne Park.
With the victory, Tiafoe, 20, advances to face No. 5 seed Kevin Anderson, who overcame a traditional poor first-round performance Sunday with a 6-3, 5-7, 6-2, 6-1 win over Adrian Mannarino, in the round of 64 on Tuesday.
Tiafoe, now ranked 39th in the world, earned the first Grand Slam match victory of his career by defeating Mikhail Kukushkin of Kazakhstan, 6-1, 6-7 (3-7), 6-3, 6-2, at the 2017 Australian Open. He fell in the opening round last year in Melbourne.
An alumnus of the Junior Tennis Champions Center in College Park, Tiafoe is coming off a 2018 season in which he secured his first ATP Tour title with a victory at the Delray Beach (Fla.) Open, becoming the youngest American to win a title since Andy Roddick won at age 19 in 2002 in Houston.
Gunneswaran qualified for his maiden Grand Slam after beating world No. 192 Yosuke Watanuki in the final round of qualifying.
Denis Kudla, a fellow JTCC alum from Ukraine, rallied past Melbourne resident Marc Polmans, a former junior doubles Australian Open champion, 5-7, 1-6, 6-2, 6-3, 6-2, early Monday.
ESPN and Tennis Channel will have full television coverage of the tournament.