Air France AF8995 Flight Status Today

Domestic flight Air France AF8995 takes off from Columbus (CMH) United States to New York (JFK) United States. It's operated by Air France. The plane leaves John Glenn Columbus International Airport at 17:42 America/New_York. The flight is expected to land at John F. Kennedy International Airport at 19:45 America/New_York. The flight will last about 1 hours 59 minutes.

AF8995 - AFR8995 Air France (AF)
CMH Columbus United States
Flight CMH to JFK
JFK New York United States
NO FLIGHTS IN THE LAST 14 DAYS Lasted record on 11-March-2020
Departure John Glenn Columbus International Airport
IATA: CMHICAO: KCMH
Scheduled 17:42
None C54
Arrival John F. Kennedy International Airport
IATA: JFKICAO: KJFK
Scheduled 19:45
2 C70
AF8995 Detail
  • Type: Domestic Flight
  • Flight Duration: 1 hours 59 minutes
  • Flight Distance: 776 kms / 482 miles
Airline
  • Air France
  • IATA: AF
  • ICAO: AFR
  • Operating Days: NO FLIGHTS IN THE LAST 14 DAYS
Passenger Services
  • Service type: Normal passenger
  • Seats: 50
  • Freight capacity: 1.5 tons
  • Passenger classes: First Class, Economy, Shuttle, Premium Economy
More Detail
  • Aircraft: CRJ
  • Callsign: AFR8995
  • Departure Timezone: America/New_York
  • Arrival Timezone: America/New_York
  • Current Time in Columbus: Sunday 2026-01-18 04:36 AM
  • Current Time in New York: Sunday 2026-01-18 04:36 AM

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Air France Flight AF8995 FAQs

What is the scheduled flight duration for Air France AF8995 flight?

On average, nonstop flight takes 1 hour(s) 59 minutes, with the flight distance of 776 km (482 miles).

What type of aircraft is used for the Air France AF8995 flight?

All Air France AF8995 flights are operated using CRJ aircraft.

Which terminal the flight Air France AF8995 is arriving at?

Flight Air France AF8995 arrives in John F. Kennedy International Airport at Terminal 2.