Jazeera Airways J9403 Flight Status Today

International flight Jazeera Airways J9403 takes off from Kuwait (KWI) Kuwait to Hyderabad (HYD) India. It's operated by Jazeera Airways. The plane leaves Kuwait International Airport at 18:50 Asia/Kuwait. The flight is expected to land at Rajiv Gandhi International Airport at 01:40 Asia/Kolkata. The flight will last about 4 hours 20 minutes.

J9403 - JZR403 Jazeera Airways (J9)
KWI Kuwait Kuwait
Flight KWI to HYD
HYD Hyderabad India
Contact Airlines It may not operate on the date requested
Departure Kuwait International Airport
IATA: KWIICAO: OKBK
Scheduled 18:50
5 None
Arrival Rajiv Gandhi International Airport
IATA: HYDICAO: VOHS
Scheduled 01:40
None None
J9403 Detail
  • Type: International Flight
  • Flight Duration: 4 hours 20 minutes
  • Flight Distance: 3375 kms / 2097 miles
Airline
  • Jazeera Airways
  • IATA: J9
  • ICAO: JZR
  • Operating Days: Tuesday, Wednesday, Saturday
Passenger Services
  • Service type: None
  • Seats: None
  • Freight capacity: None
  • Passenger classes: None
More Detail
  • Aircraft: Airbus A320
  • Callsign: F-VOHS2
  • Departure Timezone: Asia/Kuwait
  • Arrival Timezone: Asia/Kolkata
  • Current Time in Kuwait: Thursday 2026-01-01 02:13 AM
  • Current Time in Hyderabad: Thursday 2026-01-01 04:43 AM

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Jazeera Airways Flight J9403 FAQs

What is the scheduled flight duration for Jazeera Airways J9403 flight?

On average, nonstop flight takes 4 hour(s) 20 minutes, with the flight distance of 3375 km (2097 miles).

What type of aircraft is used for the Jazeera Airways J9403 flight?

All Jazeera Airways J9403 flights are operated using Airbus A320 aircraft.

How many Jazeera Airways J9403 flights are operated a week?

3 flights per week. The Flight Jazeera Airways J9403 is operated on Tuesday, Wednesday, Saturday.