A jet airplane traveling at the speed of 500 km/hr ejects its products of combustion at the speed of 1500 km/hr relative to the jet plane. What is the speed of the later with respect to an obserever on the ground? 

Asked by Topperlearning User | 4th Jun, 2014, 01:23: PM

Expert Answer:

Let Vg be the velocity of ejecting gases with respect to ground.

velocity of jet plane  Vj = + 500 km/hr

Velocity of ejected products w.r.t. jet plane

 Vgj = Vg - Vj

or,  Vg = Vgj   + Vj

            = -1500 + 500 = -1000 km/hr

 

 

Answered by  | 4th Jun, 2014, 03:23: PM