Department of Defense to conduct flyover of New York City for 4th of July

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The Department of Defense will conduct a flyover of New York City on July 4 to “recognize the role of the city in the birth of this great nation,” the Pentagon said Thursday.

Multiple aircraft from the U.S. Air Force and the U.S Marine Corps are set to fly down the Hudson River at approximately 5 p.m. on Saturday, July 4, and are expected to pass the Statue of Liberty. The aircraft will fly over New York City in five waves.

MOUNT RUSHMORE JULY 4 CELEBRATION TO FEATURE TRUMP REMARKS, FLYOVER, FIREWORKS

The flyover will be led by the U.S. Air Force Thunderbirds, followed by B-1, B-2 and B-52 bombers, F-15 and F-22 fighters and U.S. Marine Corps F-35 fighters.

The Pentagon said Thursday that they are “proud to be part of this celebration of the nation’s 244th birthday.”

“This flyover represents a small part of our military capacity and capability to continue to protect and defend the nation,” they said in a statement. “Homeland defense is our top priority.”

DETAILS OF DC'S FOURTH OF JULY CELEBRATIONS

The flyover of New York City is just one of several across the country to commemorate Independence Day.

On Friday, a military flyover will take place during an event at Mount Rushmore which will feature fireworks and remarks from President Trump. On Saturday, another military flyover will be conducted for the Washington D.C. celebrations.