Question

A jet airplane travelling at the speed of $500\text{km/h}$ ejects its products of combustion at the speed of $1500\text{km/h}$ relative to the jet plane. What is the speed of the latter with respect to an observer on the ground (in $\text{km/h}$) ?

Answer 1

Let us assume the direction of the velocity of the airplane to be negative. So, if we take the velocity of airplane to be $V_P$ and that of the products of combustion to be $V_C$, then

$V_P=-500\text{km/h}$
and relative velocity of products of combustion, $V_C-V_P=1500\text{km/h}$.
[since the combustion products are released opposite to the direction of plane's velocity]

$\Rightarrow V_C-(-500)=1500$
$\Rightarrow V_C+500=1500$
$\Rightarrow V_C=1500-500=1000\text{km/h}$
is the velocity w.r.t an observer on the ground.