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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.
Nuclear bomb damaged in crash [34] During a simulated takeoff, a wheel casting failure caused the tail of a USAF B-47 carrying a Mark 36 Mod 1 nuclear bomb to hit the runway, rupturing a fuel tank and sparking a fire which burned for some 7 hours. [35] The weapon used in-flight insertion and the weapon was in its retracted, unarmed state. [36]
Richardson dropped the bomb into the shallow waters of Wassaw Sound, near the mouth of the Savannah River, a few miles from the city of Tybee Island, where he believed the bomb would be swiftly recovered. The Pentagon recorded the incident in a top secret memo to the chairman of the United States Atomic Energy Commission (AEC).
A frightening moment in the 1950s has mostly been forgotten today — the release of an unloaded nuclear bomb by the Air Force over the Palmetto State. That time the U.S. government accidentally ...
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On February 5, 1958, during a training mission flown by a B-47, a Mk 15 nuclear bomb, was lost off the coast of Tybee Island, Georgia near Savannah. Initially, experts disputed whether the bomb was nuclear or not. If it contains a plutonium core, it is a fully-functional nuclear weapon.
In 1964, United Nuclear Corporation opened a nuclear facility in Rhode Island. A few months later, an accident happened.
Tybee Island is home to the first of what eventually became the Days Inn chain of hotels, the oft-photographed Tybee Island Light Station, and the Fort Screven Historic District. On February 5, 1958, the U.S. Air Force accidentally dropped an atomic bomb into the sea off Tybee Island due to an accidental collision between two aircraft.