A jet airplance 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 ?

Explanation

Let us consider that `v_(p)`=velocity of products
`v_(A)=` velocity of airplane
`v_(p)-v_(A)=-1500km h^(-1)`
`v_(p)-500=-1500`
`v_(p)=-1500+500`
`=-1000km h^(-1)` (negative sign indicates that velocity of products is opposite to that of jet).