February 11, 2014 Bjorn Broten 4. The bomb has … 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-47bomber carrying the bomb. Also, it involved more than the B-47, making the current title somewhat POV. 23.12.2014 - 1958 Tybee Island mid-air collision is located in Georgia (U.S. state) Home 1958 Tybee Island mid-air collision 1958 Tybee Island mid-air collision. 1958 Tybee Island mid-air collision; The United States Air Force lost a 7,600-pound Mark 15 nuclear bomb. Fifty years ago, a B-47 bomber dropped a 7,000-pound nuclear bomb into the waters off Tybee Island, Ga., after a mid-air collision. No one knows where it is (xpost from r/UnexplainedPhotos) Therefore it is mischaracterized as a crash. Midair collision of article 1958 Tybee Island mid-air collision: The B-47 bomber was on a simulated combat mission from Homestead Air Force Base in Florida . 1958 Tybee Island B-47 crash → 1958 Tybee Island mid-air collision –The B-47 landed. When a B-47 pilot got rid of a nuclear bomb near Savannah, Georgia . --Whoop whoop pull up Bitching Betty | Averted crashes 11:34, … Military Aviation . It … The Tybee Island mid-air collision was an incident on February 5, 1958, in which the United States Air Force lost a 7600 lb Mark 15 nuclear bomb in the waters off Tybee Island …