Question

A jet airplane travelling at the speed of ` 500 km ^(-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 burnt gases with respect to observer on the ground ?

Answer 1

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, `v_("smoke") = v′_("smoke") – v_("jet")– 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.