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Russia's Sharapova wins Australian Open

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Russian tennis star Maria Sharapova became the winner of the 2008 Australian Open in Melbourne after beating Serbia's Ana Ivanovic in the finals early on Saturday.
MOSCOW, January 26 (RIA Novosti) - Russian tennis star Maria Sharapova became the winner of the 2008 Australian Open in Melbourne after beating Serbia's Ana Ivanovic in the finals early on Saturday.

Sharapova, 20, the number 5 seed at the tournament, won the Saturday match in two straight sets (7:5, 6:3). The game lasted for about an hour and a half.

On the way to the finals, the former World No 1 had stormed past the current leader in the World ratings, Justine Henin of Belgium, in the quarterfinals on Tuesday with a 6-4, 6-0 straight sets victory, and made equally light work of Serbia's Jelena Jankovic, winning 6-3, 6-1 in the semifinals on Thursday.

Sharapova, beaten by Serena Williams in the Australian Open final a year ago, now has three Grand Slam titles and is lacking only the French Open title to achieve her Career Grand Slam.

The Russian player's two previous Grand Slam titles have come at the 2004 Wimbledon and the 2006 U.S. Open.

On Wednesday evening it was announced that Sharapova had been included in the Russian women's team to face Israel in a quarterfinal Federation Cup match, her debut in the national team.

The star's long-awaited appearance in the Russian side is undoubtedly connected to the fact that players who do not actively represent their country in international tournaments will be barred from taking part in the upcoming Summer Olympics, to be held in Beijing on August 8-24.

Russia are due to face Israel in the Federation Cup on February 2-3. The match will be played in the Israeli city of Ramat HaSharon.

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