How tall is Bobby Morrow?

Bobby Morrow's listed height was 186 cm (6'1"). He was taller than about 91% of men in the United States.

Where 186 cm sits among men in the United States

Taller than about 91% of men in the United States born in 1935 (average 175.9 cm).

About Bobby Morrow

Bobby Morrow was an American sprinter who won three gold medals at the 1956 Olympics. He was born on 15 October 1935 in Harlingen, Texas, and was raised on a cotton and carrot farm on the outskirts of San Benito, Texas. He played football for San Benito High School before becoming a sprinter, and later ran for Abilene Christian University.

Morrow won the 1955 AAU 100-yard title. His most successful season was 1956, when Sports Illustrated chose him as Sportsman of the Year, and he won the sprint double in the national college championships. At the 1956 Summer Olympics in Melbourne he led the American sprint team and won three gold medals, the first in the 100-meter dash. He then led an American sweep of the medals in the 200-meter dash, equaling the world record at that distance with a time of 20.6 seconds.

At a listed 186 cm (6'1"), Morrow stood that height while winning his Olympic titles. He died on 30 May 2020. He has been called the dominant sprinter of the 1950s.

Frequently asked

How tall is Bobby Morrow in feet and inches? Bobby Morrow was 6'1" tall, which is 186 cm, 73.2 inches or 1.86 m.

Is Bobby Morrow tall for men in the United States? At 186 cm he was taller than about 91% of men in the United States born in 1935, whose average height is 175.9 cm.

Sources

Heights are as publicly listed, rounded to the nearest centimetre; listed heights for public figures are often self-reported. Biographical details adapted from Wikipedia under CC BY-SA 4.0.

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