The Tybee Island B-47 crash was an incident on February 5, 1958, in which the United States Air Force lost a Mark 15 nuclear bomb in the waters off Tybee Island near Savannah, Georgia, United States. During a practice exercise, the B-47 bomber carrying the bomb collided in midair with an F-86 fighter plane. To protect the aircrew from a possible detonation in the event of a crash, the bomb was jettisoned. Following several unsuccessful searches, the bomb was presumed lost somewhere in Wassaw Sound off the shores of Tybee Island.
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| - 1958 Tybee Island mid-air collision
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| - The Tybee Island B-47 crash was an incident on February 5, 1958, in which the United States Air Force lost a Mark 15 nuclear bomb in the waters off Tybee Island near Savannah, Georgia, United States. During a practice exercise, the B-47 bomber carrying the bomb collided in midair with an F-86 fighter plane. To protect the aircrew from a possible detonation in the event of a crash, the bomb was jettisoned. Following several unsuccessful searches, the bomb was presumed lost somewhere in Wassaw Sound off the shores of Tybee Island.
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| - Tybee Island, Georgia, United States
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| - Georgia
- A Mk 15 nuclear bomb of the type lost when jettisoned after the collision
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abstract
| - The Tybee Island B-47 crash was an incident on February 5, 1958, in which the United States Air Force lost a Mark 15 nuclear bomb in the waters off Tybee Island near Savannah, Georgia, United States. During a practice exercise, the B-47 bomber carrying the bomb collided in midair with an F-86 fighter plane. To protect the aircrew from a possible detonation in the event of a crash, the bomb was jettisoned. Following several unsuccessful searches, the bomb was presumed lost somewhere in Wassaw Sound off the shores of Tybee Island.
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