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 later with respect to an observer on the ground?​

Answer 1

Answer:

It is given that the speed of products of combustion is 1500 km/h relative to the jet plane. From the figure we know that the velocity of →vB−→vA will be in the negative direction. Hence, →vB/A=→vB−→vA=−1500km/h. This means that the speed of the products is 1000km/h with respect to an stationary observer on the ground.

Answer 2

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It is given that the speed of products of combustion is 1500 km/h relative to the jet plane.

From the figure we know that the velocity of →vB−→vA will be in the negative direction.

Hence, →vB/A=→vB−→vA=−1500km/h.

This means that the speed of the products is 1000km/h with respect to an stationary observer on the ground.

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