Share

COLUMBUS, Ohio – Ohio State senior and Upper Arlington, Ohio, native Kirsten Flower has been named the Big Ten Women’s Tennis Athlete of the Week, the Big Ten Conference announced Tuesday. Flower helped lead the 33rd-ranked Buckeyes to three team victories over Top 50 opponents last week, with a home win over No. 34 Kentucky and road victories over No. 50 William & Mary and No. 23 Virginia. She was 3-0 at No. 1 singles for the Buckeyes without dropping a set, and clinched the team victory in two matches.

Against the Wildcats, Flower defeated Megan Broderick, 6-0, 6-3, en route to a 7-0 Buckeye shutout win. In the 5-2 victory over the Tribe, Flower posted a straight sets win against Ragini Acharya, 6-2, 6-2. Flower then downed No. 72 Lindsey Hardenbergh of the Cavaliers in straight sets, 7-6, 6-4, to clinch the 5-2 upset for the Buckeyes.

The weekly laurel is the second for Flower and the first for Ohio State since her first in March 2009.
 
Flower has been a stalwart at the No. 1 position for the Buckeyes since transferring from Georgia Tech following her sophomore year. Last season she led the team in singles wins (19) on the way to an All-Big Ten selection and NCAA Tournament appearance for the Scarlet and Gray. This year she has jumped out to a 10-4 start, including a 7-2 ledger in dual play to aid the Buckeyes to an 8-1 record.

After a week off, Flower and the Buckeyes return to the courts with matches against LSU at 2 p.m. Feb. 17 and against Tennessee at noon Feb. 20. Admission is free to all Ohio State regular-season matches, which are held in the Ohio State University Varsity Tennis Center.