3.5 A jet airplane travelling at the speed of 500 km h–1 ejects its products of combustion at the speed of 1500 km h–1 relative to the jet plane. What is the speed of the latter with respect to an observer on the ground?

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8 months ago

3.5 The speed of the jet plane, Vj = 500 kmph. If the speed of the product of combustion is Vpc, then V pc = Vpc-Vj = -1500 kmph (relative to the jet plane)

Speed of the combustion product relative to the observer on the ground,

Vpc = -1500 + Vj = -1500+500 = -1000 kmph

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