A jet airplane travelling at the speed of 500 km/h ejects its products of combustion at the speed of 1500 km/h relative to the jet plane. What is the speed of the latter with respect to an observer on the ground?
Given,
Speed of the jet plane, vjet = 500 km/h
Relative speed of the combustion products, v’smoke = 1500km/h
Observation:
As plane moves forward, combustion smoke moves backward.
Thus,
∴ Velocity of combustion gases, vsmoke = vjet – v’smoke
= 500 – 1500
= -1000 m/s
(∵ Jet moving direction is taken as ‘+ve’ for man standing on ground)