Question

A jet aeroplane travelling at the speed of 500 kmh^-1, ejects its products of combustion at the speed of 1500 km^-1 relative to the jet plane. What is the speed of the latter with respect to an observer on the ground?

Answer 1

Velocity of jet = v_j = 500 kmh^-1 (away from the observer)

Relative velocity of ejection with respect to jet = v_cj = 1500 kmh^-1

If 'v_c' is the velocity of the ejection (of ejected products), then 

V_ej = v_e - e_j

or v_e = v_ej + v_j = 1500 + (-500) = 1000 kmh^-1.