 A Mk 15 nuclear bomb of the type lost when jettisoned after the collision | |
| Accident | |
|---|---|
| Date | February 5, 1958 |
| Summary | Mid-air collision |
| Site | Tybee Island, Georgia, U.S. 32°0′N 80°51′W / 32.000°N 80.850°W |
| First aircraft | |
| Type | Boeing B-47B Stratojet |
| Operator | United States Air Force (Strategic Air Command) |
| Registration | 51-2349 |
| Crew | 3 |
| Fatalities | 0 |
| Second aircraft | |
| Type | North American F-86L Sabre |
| Operator | United States Air Force (Tactical Air Command) |
| Registration | 52-10108[1] |
| Crew | 1 |
| Fatalities | 0 |
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 off 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.
The bomb was jettisoned to help prevent a crash and explosion. After several unsuccessful searches, the weapon was declared lost in Wassaw Sound off the shores of Tybee Island.