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A jet airplane travelling at the speed of 500 km h-1 ejects its products of combustion at the speed of 1500 km h-1 relative to the jet plane. What is the speed of the latter with respect to an observer on the ground?
Asked by Abhay Last Modified
Answer 
Both are in opposite direction,so consider one velocity in negative direction.
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 sign means downward .
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