A jet airplane travelling at the speed of 500 km h-1 ejects its products of combustion at the speed of 1500 km h-1 relative to the jet plane. What is the speed of the latter with respect to an observer on the ground ?
Answers
Speed of the jet airplane, vjet = 500 km/h
Relative speed of its products of combustion with respect to the plane,
vsmoke = – 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.
Solution:
Speed of the jet airplane, vjet = 500 km/h
Relative speed of its products of combustion with respect to the plane,
vsmoke = – 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.