Answer :

Given,

Speed of the jet plane, vjet = 500 km/h

Relative speed of the combustion products, v’smoke = 1500km/h

Observation:

As the plane moves forward, combustion smoke moves backwards.

Thus,

Velocity of combustion gases, vsmoke = vjet – v’smoke

= 500 – 1500

vsmoke = -1000 km/h

Thus, the velocity of gases will appear leaving the jet at a speed of 1000 km/h

( Jet moving direction is taken as ‘+ve’ for the man standing on the ground)

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