Kalitta Air K49707 Flight Status Today

Domestic flight Kalitta Air K49707 takes off from New York (JFK) United States to Sumter (SSC) United States. It's operated by Kalitta Air. The plane leaves John F. Kennedy International Airport at 11:01 America/New_York. The flight is expected to land at Shaw Air Force Base at 12:30 America/New_York. The flight will last about 1 hours 2 minutes.

K49707 - CKS9707 Kalitta Air (K4)
JFK New York United States
Flight JFK to SSC
SSC Sumter United States
Contact Airlines It may not operate on the date requested
Departure John F. Kennedy International Airport
IATA: JFKICAO: KJFK
Scheduled 11:01
None None
Arrival Shaw Air Force Base
IATA: SSCICAO: KSSC
Scheduled 12:30
None None
K49707 Detail
  • Type: Domestic Flight
  • Flight Duration: 1 hours 2 minutes
  • Flight Distance: 947 kms / 589 miles
Airline
  • Kalitta Air
  • IATA: K4
  • ICAO: CKS
  • Operating Days: Tuesday
Passenger Services
  • Service type: None
  • Seats: None
  • Freight capacity: None
  • Passenger classes: None
More Detail
  • Aircraft: Boeing 747-4B5(BCF)
  • Callsign: T-F5M
  • Departure Timezone: America/New_York
  • Arrival Timezone: America/New_York
  • Current Time in New York: Sunday 2026-01-18 12:58 PM
  • Current Time in Sumter: Sunday 2026-01-18 12:58 PM

Kalitta Air Flight K49707 FAQs

What is the scheduled flight duration for Kalitta Air K49707 flight?

On average, nonstop flight takes 1 hour(s) 2 minutes, with the flight distance of 947 km (589 miles).

What type of aircraft is used for the Kalitta Air K49707 flight?

All Kalitta Air K49707 flights are operated using Boeing 747-4B5(BCF) aircraft.

How many Kalitta Air K49707 flights are operated a week?

1 flights per week. The Flight Kalitta Air K49707 is operated on Tuesday.