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. The speed of the products of combustion with respect to an observer on the ground is:

• A. $500km{h}^{-1}$
• B. $1000km{h}^{-1}$
• C. $1500km{h}^{-1}$
• D. $2000km{h}^{-1}$

Answer 1

The correct option is B $1000km{h}^{-1}$

Velocity of jet plane w.r.t. ground
$V_{JG}=500km{h}^{-1}$
Velocity of products of combustion w.r.t. jet plane

$V_{CJ}=1500km{h}^{-1}$
$\therefore$ Velocity of products of combustion w.r.t. ground is
$v_{CG}=v_{CJ}+v_{JG}=-1500km{h}^{-1}+500km{h}^{-1}=-1000km{h}^{-1}$
-ve sign shows that the direction of products of combustion is opposite to that of the plane.
$\therefore$ Speed of the products of combustion w.r.t. ground =$1000km{h}^{-1}$