Jankovic outguns Kuznetsova
Kuznetsova won the US Open in 2004
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Monday, 04, Sep 2006 09:46
Serbia's Jelena Jankovic has reached her first grand slam quarter-final after beating 2004 US Open champion Svetlana Kuznetsova.
The 21-year-old lost the first set on a tie-break but came back to power past the Russian 6-7 (5-7) 6-3 6-2.
The 19th seed had been in good form coming into the tournament at Flushing Meadows and Kuznetsova is her second major scalp, after disposing of ninth seed Nicole Vaidisova in the last round.
Jankovic will now play another Russian, Elena Dementieva, after the fourth seed beat French teenager Aravane Rezai 7-5 6-4.
The 2004 runner-up has yet to drop a set at the 2006 tournament and managed to contain the 19-year-old's powerful strokes.
Second seed Justine Henin-Hardenne looked very impressive as she destroyed Israel's Shahar Peer 6-0 6-1.
The 2003 winner had to survive a tough three-set evaluation from 14th seed Francesca Schiavone in the last round but looked in imperious form as she won in just 50 minutes on the Arthur Ashe court.
The Belgian will now play American Lindsay Davenport after she saw off the challenge of seventh seed Patty Schnyder in straight sets.
The 1998 champion has missed much of this season because of injury but ousted the Swiss player 6-4 6-4 to reach the last eight.