Question

A jet airplane traveling 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 ejection with respect to an observer on the ground?

Answer 1

velocity of plane = 500
velocity of ejected products with respect to plane = 1500
=> Vejected products - Vplane= 1500

=>Vejected products = Vplane+ 1500 = 2000km/h

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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