Question

A jet plane travelling at a speed of 500 km/hr ejects its products of combustion at the speed of 1500km/hr relative to the jet plane.what is the speed of the later with respect to the observer on the ground

Answer 1

Clearly, the direction of jet and that of products of combustion is opposite.
So, in order to find speed of products of combustion with respect to stationary objet ( observer on ground) , we need to find the difference in their speeds.
Speed = 1500-500=1000 km/hr

Answer 2

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Speed of the jet airplane, v jet = 500 km/h

Relative speed of its products of combustion with respect to the plane,

v smoke = – 1500 km/h

Speed of its products of combustion with respect to the ground = v′smoke

Relative speed of its products of combustion with respect to the airplane,

vsmoke = v′smoke – vjet

1500 = v′smoke – 500

v′smoke = – 1000 km/h

The negative sign indicates that the direction of its products of combustion is opposite to the direction of motion of the jet airplane.

I hope, this will help you

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