A jet airplane travelling at a 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?

The correct option is B $1000\text{km/h}$

Let $v_j{v}_g$ and $v_0$ be the velocities of jet, ejected gases and observer on the ground respectively.
Let jet be moving towards right (+ve direction.)
$\therefore$ Ejected gases will move towards left (-ve direction).
$\therefore$ According to the statement
$v_j-v_0=500\text{km/h}\dots \dots \left(i\right)$
As observer is at rest
$v_g-v_j=-1500\text{km/h}\text{(given)}\dots \dots \left(i\right)$
$\therefore$ Adding Eqs. (i) and (ii), we get the speed of combustion products w.r.t., observer on the ground
$(v_j-v_0)+(v_g-v_j)=v_g-v_0=500+(-1500)$
Or $v_g-v_0=-1000\text{km/h}$
-ve sign shows that relative velocity of the ejected gases w.r.t. observer is towards left or in a direction opposite to the motion of the jet plane.