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 the ground ?

Answer 1

$\overrightarrow{V_j}$ is the velocity of jet

${\overrightarrow{V}}_{cj}$ the velocity of combustion product with respect to jet

${\overrightarrow{V}}_{cj}={\overrightarrow{V}}_c-{\overrightarrow{V}}_j$ where

${\overrightarrow{V}}_c$ is the velocity of combustion products with respect to earth.

${\overrightarrow{V}}_c={\overrightarrow{V}}_{cj}-{\overrightarrow{V}}_j$

We take the direction of jet propagation as positive direction

${\overrightarrow{V}}_c$=-$1500+500=-1000km/h$

Velocity of combustion products with respect to observer on ground is 1000km/h