a jet airplane is travelling at a speed of 500km/h.it ejects its products at the speed of 1500km/h relative to the jet plane.what is the speed of the latter with respect to an observer at the ground?

Dear Abhishek Choudhary,

      Velocity of jet w.r.t. ground V_jg= 500 km/h                                                                …...(Upward)

        Velocity of products relative to jet V_PJ = 1500 km/h                                                …(downward)

Hence, velocity of  products relative to ground = -1500+500

                                                                                    =  -1000 km/h

Here – ve sing means downward .

 

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Mr. Ali

IIT Delhi

Dear Student,
Please find below the solution to your problem.

Velocity of jet w.r.t. ground Vjg= 500 km/h …...(Upward)
Velocity of products relative to jet VPJ = 1500 km/h ……(downward)
Hence, velocity of products relative to ground = -1500+500 = -1000 km/h
Here – ve sing means downward .

Thanks and Regards