Question

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?

Answer 1

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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,

v(smoke) = v′(smoke) – v(jet)

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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