A Jet airplane traveling at the speed of 500km/hr ejects its products of combustion at the speed of 1500km/h relative to the Jet plane. What is the speed of the latter with respect to an observer on the ground?
CLASS - XI PHYSICS (Kinematics)
Answers
Velocity of latter wrt jet= -1500km/h (-ve sign indicates the motion in opposite direction)
Velocity of latter w.r.t groung = 500+(-1500)
=-1000km/h
-ve sign shows opposite direction
Hence required Velocity is 1000km/h
1000 km/hr
Given:
Velocity of jet = 500 km/hr (w.r.t to the observer)
Velocity of the observer = 0 km/hr
Velocity of combustion of products with respect to the plane = 1500 km/hr
To Find:
The speed of the latter with respect to an observer on the ground.
Calculating:
Velocity of Jet - Velocity of Observer
= 500 - 0
= 500 km hr^-1
Velocity of gas with respect to plane = velocity of gasses with respect to observer - velocity of jet with respect o observer
Substituting all the values known to us in this formula we get:
Vg/p = vg - (-500)
we take - 500 because it is a vector quantity and the fuel is getting pushed in the opposite direction.
1500 = vg + 500
vg = 1500 - 500
vg = 1000 km/hr
Therefore, the speed of the later with respect to an observer on the ground is 1000 km/hr.