AUCKLAND, New Zealand (AP)—Top-seeded David Ferrer won the Heineken Open for the third time, beating unseeded Belgium’s Olivier Rochus 6-3, 6-4 on Saturday.
Ferrer, ranked fifth in the world, completed preparations for the Australian Open with an impressive performance, dropping only one set over four matches for his 12th career title.
He also won the event in 2007 and last year.
Ferrer beat Lukas Rosol, Alejandro Falla and third-seeded Fernando Verdasco on the way to the final, then overpowered the 68th-ranked Rochus with his powerful serve and forehand. Ferrer had nine aces and won 81 percent of first-serve points.
Ferrer, ranked fifth in the world, completed preparations for the Australian Open with an impressive performance, dropping only one set over four matches for his 12th career title.
He also won the event in 2007 and last year.
Ferrer beat Lukas Rosol, Alejandro Falla and third-seeded Fernando Verdasco on the way to the final, then overpowered the 68th-ranked Rochus with his powerful serve and forehand. Ferrer had nine aces and won 81 percent of first-serve points.
“I feel tired, very tired. I ran a lot this week,” Ferrer said. “But it’s the first time I’ve won three times in the same tournament and so this is my favorite tournament. For me to win a tournament three times is very special.”
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