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 ground?
Answer:
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
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.
No comments:
Post a Comment