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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 kmh-1relative to the jet plane. What is the speed of the latter with respect to an observer on the ground?

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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 ...
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Both are in opposite direction,so consider one velocity in negative direction. Velocity of jet w.r.t. groundVjg= 500 km/h …...(Upward) Velocity of products relative to jetVPJ= 1500 km/h …(downward) Hence, velocity of products relative to ground = -1500+500 = -1000 km/h Here – ve sign means downward . read less
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here I have attatched the image of solution.
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