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प्रश्न
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?
उत्तर १
Speed of the jet airplane, vjet = 500 km/h
Relative speed of its products of combustion with respect to the plane,
vsmoke = – 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,
vsmoke = v′smoke – vjet
– 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.
उत्तर २
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.
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