Question

A jet air plain traveling at a speed of 500 km/hr ejects its products of combustion at the speed of 1500 km/hr relative to the jet plain What is the speed of the latter with respect to an observer on the ground

Answer 1

Hi.....

Velocity of jet w.r.t. ground Vjg= 500 km/h …...(Upward)

Velocity of products relative to jet VPJ = 1500 km/h …(downward)

Hence, velocity of products relative to ground = -1500+500

= -1000 km/h

Here – ve sing means downward ...

Hope this helps u!!

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,

vsmoke = v′smoke – vjet

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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