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A jet airplane travelling at the speed of 500km{{h}^{-1}} ejects its product of combustion at the speed of 1500km{{h}^{-1}} relative to the jet plane. The speed of the products of combustion with respect to an observer on the ground is

Option: 1

500km{{h}^{-1}}


Option: 2

1000km{{h}^{-1}}


Option: 3

1500km{{h}^{-1}}


Option: 4

2000km{{h}^{-1}}


Answers (1)

best_answer

Velocity of jet plane with respect to ground,

{{v}_{JG}}=500km{{h}^{-1}}

Velocity of products of combustion with respect to jet plane

{{v}_{CJ}}=-1500km{{h}^{-1}}

\therefore Velocity of products of combustion with respect to ground is

{{v}_{CG}}={{v}_{CJ}}+{{v}_{JG}}=-1500km{{h}^{-1}}+500km{{h}^{-1}}=-1000km{{h}^{-1}}

-ve sign shows that the direction of products of combustion is opposite to that of the plane.

\therefore Speed of the products of combustion with respect to ground = 1000km{{h}^{-1}}

Posted by

Irshad Anwar

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