^{–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 ground?

Hi,

Here we will be considering the direction of velocity of the jet to be positive direction.

Given:

Speed of the jet w.r.t ground = 500 km/hr (t

Speed of the combustion product w.r.t jet =- 1500 km/hr (negative sign because it is in the direction opposite to the direction of velocity of the jet)

Therefore, applying the concept of relative velocity we get

Negative sign indicates that the velocity of the gas is in the direction opposite to the direction of the velocity of the jet.

Hope this helps,

Good luck.

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