How High Is Seymour Mountain?

Seymour Mountain is 1,256 meters (4,120 feet) tall and is located in New York in United States of America. Seymour Mountain has a prominence of 311 meters (1,020 feet).

Height of Seymour Mountain in meters: 1,256 meters

Height of Seymour Mountain in feet: 4,120 feet

Country: United States of America

Prominence of Seymour Mountain in meters: 311 meters

Prominence of Seymour Mountain in feet: 1,020 feet

Read next: How High Is Phillips County High Point?

You might also like: The 10 Highest Mountains in United States of America