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


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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.

Explanation:

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Answer 2

QUESTION:-

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

EXPLANATION:-

Given:-

$\bf\;V_{(airplane)}=500\;km/hr$

speed of combustion=$\bf\;=v_{smoke}=-v_{jet}$

So,

$\bf\;v_{(smoke)}=v'_{(smoke)}-v_{(jet)}$

=>v_(smoke)=-1500

Putting values;-

-1500=v'-500

v'=-1000 km/hr