Answer :

According to the question, the speed of the jet airplane (V_A) is 500 km/h

And the speed of ejection of combustion products relative to the jet plane is given by

V_B – V_A= – 1500 km/h

Here, the negative symbol denotes that the combustion products are moving in the opposite direction of the jet.

And V_B is the speed of ejection of combustion products with respect to the observer on the ground. It is given by

V_B – 500 = – 1500

Or, V_B = – 1500 + 500

Therefore, V_B = – 1000 km/h