Question

A jet airplane travelling at the speed of 500 km / hr ejects its products of combustion at the speed of 1500 km / hr relative to the jet plane. What is the speed of the latter with respect to an observer on the ground?​

Answer 1

Speed of ejection of combustion products observed from ground is 1000km/h in direction opposite to the direction of motion of the jet airplane.

Answer 2

given:

`v{j} = 500`

`vpj = - 1500`

solution:

`vpj = vp - vj`

`- 1500 = vp - 500`

`vp = - 1000`

= 1000 km/h