1958 Tybee Island mid-air collision

When

5 Feb 1958

Kind

Air disaster

Coordinates

32.00° N, 80.85° W

Description

aircraft collision that involved the loss of a nuclear bomb

Deaths

0

US Air Force incident involving a nuclear bomb

The Tybee Island mid-air collision was an incident on February 5, 1958, in which the United States Air Force lost a 7,600-pound (3,400 kg) Mark 15 nuclear bomb in the waters of Tybee Island near Savannah, Georgia, United States. During a night practice exercise, an F-86 fighter plane collided with the B-47 bomber carrying the large weapon.

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• Wikidata: air, rail, road, mining and industrial accidentsWikidata contributorsCreative Commons CC0 1.0 (public domain dedication)Source page
• Wikipedia year pages (English), by Wikipedia contributorsWikipedia contributors · names it “1958 Tybee Island mid-air collision: A U.S. B-47 bomber jettisons a hydrogen bomb into Wassaw Sound off Tybee Island, Georgia; it is never recovered”Creative Commons Attribution-ShareAlike 4.0Source page
• Wikipedia year pages (German), by Wikipedia contributorsWikipedia contributors · names it “Nach einem Zusammenstoß mit einem North American F-86-Jagdflugzeug in der Luft muss die Crew eines beschädigten Boeing B-47-Bombers eine Wasserstoffbombe ohne…”Creative Commons Attribution-ShareAlike 4.0Source page
• Wikipedia year pages (Russian), by Wikipedia contributorsWikipedia contributors · names it “В США в результате авиационного происшествия над побережьем штата Джорджия бомбардировщик ВВС США аварийно сбросил в океан водородную бомбу Mark 15.”Creative Commons Attribution-ShareAlike 4.0Source page