A jet airplane travelling at the speed of $500 \mathrm{~km} \mathrm{~h}^{-1}$ ejects its products of

Question

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

Professor PC Thomas Classes · Answer

(b) : Velocity of jet plane w.r.t. ground $v_{J G}=500 \mathrm{~km} \mathrm{~h}^{-1}$ Velocity of products of combustion w.r.t. jet plane $v_{C J}=-1500 \mathrm{~km} \mathrm{~h}^{-1}$ ∴ Velocity of products of combustion w.r.t. ground is $v_{C G}=v_{C I}+v_{J G}=-1500 \mathrm{~km} \mathrm{~h}^{-1}+500 \mathrm{~km} \mathrm{~h}^{-1}=-1000 \mathrm{~km} \mathrm{~h}^{-1}$ -ve sign shows that the direction of products of combustion is opposite to that of the plane. $	herefore \quad$ Speed of the products of combustion w.r.t. ground $=1000 \mathrm{~km} \mathrm{~h}^{-1}$.