Question

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

Answer 1

Speed of the jet airplane, $v_{jet}=500km/h$
Relative speed of its products of combustion with respect to the plane, $v_{smoke}=–1500km/h$
Speed of its products of combustion with respect to the ground $=v{\text{'}}_{smoke}$
Now, $v_{smoke}^{\prime }=v_{smoke}+v_{jet}=-1500+500=-1000km/h$
The negative sign indicates that the direction of its products of combustion is opposite to the direction of motion of the jet airplane.