Question

A jet airplane travelling at the speed of $500\, km \,h ^{-1}$ ejects its products of combustion at the speed of $1500\, kmh ^{-1}$ relative to the jet plane. The speed of the products of combustion with respect to an observer on the ground is

• A. $500\, kmh ^{-1}$
• B. $1000\, kmh ^{-1}$
• C. $1500\, kmh ^{-1}$
• D. $2000\, kmh ^{-1}$

Answer 1

Answer: (B)

$v_{C J}=-1500\, kmh ^{-1}=v_{C}-v_{J}$
or $v_{C}=v_{C J}+v_{J}=\left(-1500 \, kmh ^{-1}\right)+500\, kmh ^{-1}$
$=-1000\, kmh ^{-1}$