Botic van de Zandschulp has convincingly secured a place in the second round of the Australian Open. The best tennis player in the Netherlands, the number 57 in the world, defeated the German Jan-Lennard Struff in three sets: 6-4 6-3 6-2. Struff (31) is five places higher in the world ranking.
Van de Zandschulp started the duel with the German on windy track 17 with some hesitation, who only won a game once in Melbourne. He trailed 4-2, but went on to win four games in a row. Struff ended the set, in which both were still searching, with a double fault.
In the second set Van de Zandschulp came into the game better, while Struff started to make more and more mistakes. It was 4-1 for the Dutchman, who after winning the first set never ran into problems and walked away remarkably easily from the somewhat resigned Struff.
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Perhaps illustrative of the game was the way in which Van de Zandschulp took a 3-2 lead in the third set, with a break. After a strong run, Struff had the ball for the smash, but the wind made it difficult for him. Van de Zandschulp managed to recover the smash and placed the decisive break. In the last games the resistance was broken and Van de Zandschulp finished it off with a love game, after a series of misses by Struff. The third set lasted only 32 minutes and the entire match an hour and 49 minutes.
Van de Zandschulp used half of his breakpoints: five out of ten. He allowed the German only one break and hit eight aces.
Memorable Series
For 26-year-old Van de Zandschulp, it was his first Grand Slam match since his memorable streak at the US Open in September of last year. He had his breakthrough in New York by reaching the quarterfinals.
In the second round, Van de Zandschulp will face the winner of the duel between Frenchmen Ugo Humbert and Richard Gasquet. Humbert is placed 29th. Last year, Van de Zandschulp got stuck in the opening round on his debut.
On Monday, Tallon groenpoor already reached the second round of the Australian Open. He left the Italian Fabio Fognini no chance.
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