Let $v_j\ ,\ v_g$ and $v_0$ be the velocities of jet, ejected gases i.e. combustionproducts and observer on the ground respectively. Let jet be moving towards right (+ve direction). ∴ Ejected gases will move towards left (–ve direction). According to the statement $v_{j}=500\ \text{km}\ \text{h}^{-1}$ As observer is at ground i.e. at rest $\therefore\ v_0 = 0$ Now relative velocity of plane w.r.t. the observer Relative velocity of the combustion products w.r.t. jet plane –ve sign indicates that the combustion products move in a directionopposite to that of jet. ∴ Adding equations (i) and (ii), we get the speed of combustionproducts w.r.t. observer on the ground i.e.$\ (v_{j}-v_{0}\ )+\ (v_{g}-v_{j}\ )$$=v_{g}-v_{0}$$=500+(-1500)$ or $v_{g}-v_{0}$$=-1000\ \text{km}\ \text{h}^{-1}$–ve sign shows that relative velocity of the ejected gases w.r.t. observer istowards left i.e. –ve direction i.e. in a direction opposite to the motion ofthe jet plane.