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CBSE Class 11-science Answered

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 | 04 Jun, 2014, 01:23: PM
answered-by-expert 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 | 04 Jun, 2014, 03:23: PM
CBSE 11-science - Physics
Asked by dibyanshubehera2727 | 02 Jul, 2021, 07:28: AM

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