Latitude and longitude of 1958 Tybee Island mid-air collision
- Nearby Wilmington Island, United States
Satellite map of 1958 Tybee Island mid-air collision
The Tybee Island B-47 crash 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 off Tybee Island near Savannah, Georgia, United States. During a practice exercise, an F-86 fighter plane collided with the B-47 bomber carrying the bomb. To protect the aircrew from a possible detonation in the event of a crash, the bomb was jettisoned.
Latitude: 32° 00' 0.00" N
Longitude: -80° 50' 59.99" W
Nearest city to this article: Tybee Island
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