Ryan Rowe (born April 3, 1986) is an American former professional tennis player.[1][2]
Country (sports) | |
---|---|
Born | April 3, 1986 |
Height | 6 ft 5 in (196 cm) |
Plays | Left-handed |
Prize money | $14,156 |
Singles | |
Highest ranking | No. 726 (October 8, 2012) |
Doubles | |
Highest ranking | No. 489 (July 22, 2013) |
A left-handed player from Moline, Illinois, Rowe started playing tennis at the age of nine and spent part of his childhood in Florida under the tutelage of his uncle Chris Hunt. He ranked number one in the country in the 16s and under division.[3]
Rowe played collegiate tennis for the University of Illinois and won the 2006 NCAA Division I doubles championship, partnering Kevin Anderson. In 2007, seeking to become the first pair to win back to back championships, Rowe and Anderson held a match point in the tournament decider, which they lost to Marco Born and Andreas Siljeström.[4]
Following college, Rowe competed on the professional tour, winning one singles and five ITF Futures doubles titles.[3]
ITF Futures titles
editSingles: (1)
editNo. | Date | Tournament | Surface | Opponent | Score |
---|---|---|---|---|---|
1. | Aug 2012 | USA F22, Decatur | Hard | 6–4, 3–6, 6–4 |
Doubles: (5)
editNo. | Date | Tournament | Surface | Partner | Opponents | Score |
---|---|---|---|---|---|---|
1. | Oct 2008 | USA F25, Laguna Niguel | Hard | 6–4, 6–74, [10–6] | ||
2. | Jul 2012 | USA F21, Godrey | Hard | 6–4, 6–4 | ||
3. | Aug 2012 | USA F23, Edwardsville | Hard | 6–3, 7–5 | ||
4. | Oct 2012 | USA F28, Mansfield | Hard | 6–2, 6–75, [10–7] | ||
5. | May 2013 | USA F13, Tampa | Clay | 6–3, 5–7, [12–10] |
References
edit- ^ "Quarterfinal takes on Illini flavor". Herald & Review. August 1, 2008.
- ^ "Rowe makes Beck singles final". Herald & Review. August 5, 2012.
- ^ a b "Rowe is big in stature on and off the court". Edwardsville Intelligencer. August 10, 2012.
- ^ Atlas, Brian (October 11, 2007). "All-American returns to tennis court at top of his game, owing success to friends and family". The Daily Illini.