Amapola HP6258 Flight Status : Live Tracking & Updates

Domestic flight Amapola HP6258 takes off from Houston (IAH) United States to Cleveland (CLE) United States. It's operated by Amapola. The plane leaves George Bush Intercontinental Airport at 10:45 America/Chicago. The flight is expected to land at Cleveland Hopkins International Airport at 14:11 America/New_York. The flight will last about 2 hours 26 minutes.

HP6258 - APF6258 Amapola (HP)
IAH Houston United States
Flight IAH to CLE
CLE Cleveland United States
NO FLIGHTS IN THE LAST 14 DAYS Lasted record on
Departure George Bush Intercontinental Airport
IATA: IAHICAO: KIAH
Scheduled 10:45
None None
Arrival Cleveland Hopkins International Airport
IATA: CLEICAO: KCLE
Scheduled 14:11
None None
HP6258 Detail
  • Type: Domestic Flight
  • Flight Duration: 2 hours 26 minutes
  • Flight Distance: 1757 kms / 1092 miles
Airline
  • Amapola
  • IATA: HP
  • ICAO: APF
  • Operating Days: NO FLIGHTS IN THE LAST 14 DAYS
Passenger Services
  • Service type: None
  • Seats: None
  • Freight capacity: None
  • Passenger classes: None
More Detail
  • Aircraft: None
  • Callsign: None
  • Departure Timezone: America/Chicago
  • Arrival Timezone: America/New_York
  • Current Time in Houston: Friday 2026-05-08 05:27 AM
  • Current Time in Cleveland: Friday 2026-05-08 06:27 AM

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Amapola Flight HP6258 FAQs

What is the scheduled flight duration for Amapola HP6258 flight?

On average, nonstop flight takes 2 hour(s) 26 minutes, with the flight distance of 1757 km (1092 miles).