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When the jet airplane ejects its products of combustion, we have to consider the relative velocities to find the speed of the latter with respect to an observer on the ground. Here, the speed of the jet airplane is 500 km/h, and the speed of the ejected combustion products relative to the jet plane is 1500 km/h.
To find the speed of the combustion products relative to the ground observer, we need to apply the principle of relative velocity. We add the velocity of the jet airplane to the velocity of the ejected combustion products to find their combined velocity relative to the ground.
So, the speed of the combustion products relative to the ground observer would be: Speedground=Speedjet+Speedcombustion relative to jetSpeedground=Speedjet+Speedcombustion relative to jet
Speedground=500 km/h+1500 km/h=2000 km/hSpeedground=500 km/h+1500 km/h=2000 km/h
Therefore, the speed of the combustion products with respect to an observer on the ground would be 2000 km/h.
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