Botic van de Zandschulp Height: How Tall is Botic van de Zandschulp?
Botic van de Zandschulp is a Dutch professional tennis player who was born on October 4, 1995. The Association of Tennis Professionals (ATP) gave Van de Zandschulp a career-high singles ranking of 25 on June 27, 2022.
He is now the top male singles player in the Netherlands. Additionally, he holds a world No. 222 doubles ranking, first attained on June 20, 2022, which is a career-high.
In Hamburg and Alphen, van de Zandschulp won ATP Challenger singles and doubles championships, respectively. At the 2021 Australian Open, Van de Zandschulp defeated fellow qualifier Carlos Alcaraz to earn his place in the main singles draw of a Grand Slam competition for the first time.
Van de Zandschulp began the year by defeating Adrian Mannarino and eighth-seeded Mackenzie McDonald in straight sets at the Melbourne Summer Set, where he advanced to the quarterfinals.
Despite holding match points on serve in the third set, he was defeated by Grigor Dimitrov in the quarterfinals.
At the Australian Open in 2022, he advanced to the third round before falling to World No. 2 Daniil Medvedev. He consequently debuted in the top 50 of the rankings on January 31, 2022.
How Tall is Botic van de Zandschulp?
Botic Van de Zandschulp has a height of 1.91 (6 ft 3 in). The well-known tennis player from the Netherlands weighs 85 kg (187 Ibs).