Question

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

Answer 1

Speed of the jet airplane, v_jet = 500 km/h

Relative speed of its products of combustion with respect to the plane,

v_smoke = – 1500 km/h

Speed of its products of combustion with respect to the ground = v′_smoke

Relative speed of its products of combustion with respect to the airplane,

v_smoke = v′_smoke – v_jet

– 1500 = v′_smoke – 500

v′_smoke = – 1000 km/h

The negative sign indicates that the direction of its products of combustion is opposite to the direction of motion of the jet airplane.

Answer 2

Velocity of jet airplane w.r.t observer on ground = 500 km/h.

If Vj and v0 represent the velocities of jet and observer respectively, then vj – vo = 500 km h-1

Similarly, if vc represents the velocity of the combustion products w.r.t jet plane, then vc – vg = -1500 km/h

The negative sign indicates that the combustion products move in a direction opposite to that of jet.

Speed of combustion products w.r.t. observer

= vc – u0 = (vc – vj) + (vj – v0) = (-1500 + 500) km h-1 = -1000 km h-1.