A jet airplane travelling at a speed of 500 km/h ejects its products of combustion at the speed of 1500 km/h relative to the jet plane. What is the speed of the latter with respect to an observer on the ground?

Let jet airplane be moving upwards right (+ve direction) with velocity $v_j$  and ejected gases be moving downwards (-ve direction) with velocity $v_g$while observer be at rest on the ground i.e., $v_0=0$

$\therefore v_j=500km/h$

$v_g=-1500km/h$

$v_0=0$

Relative velocity of plane with respect to the observer

$v_j-v_0=500-0=500km/h\mathrm{..............}\left(i\right)$

Relative velocity of products of combustion with respect to the jet plane

$v_g-v_j=-1500km/h\mathrm{..............}\left(ii\right)$

(velocity of ejected gas $v_g$ and velocity of $v_j$ are in opposite directions)

Adding Eqs (i) and (ii), we get

$(v_j-v_0)+(v_g-v_j)=500-1500$

$v_g-v_0=-1000km/h$

Therefore, relative velocity of the ejected gases with respect to the observer is 1000 km/h, -ve sign shows that this velocity is in a direction opposite to the motion of the jet airplane.