Let vj ,vg and v0 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
vj=500 km h−1
As observer is at ground i.e. at rest
∴ v0=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.
(vj−v0 )+ (vg−vj )
=vg−v0
=500+(−1500)
or vg−v0
=−1000 km 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.