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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. The speed of the products of combustion with respect to an observer on the ground is

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Let `vecv _(f) and vecv _(c)` be the velocities of the plane and the combustion product relative to ground respectively. `vecv _(cj)` is the velocity of combustion product relative to the jet plane.
Then `vecv _(cj) = vecv _(c ) + vecv _(i)`
or `vecv _(c) = vecv _(cj) - vec v _(j)`
taking velocity towards the observer on the ground as positive and assuming that the jet plane is flyig away from the abserver, we have, `and vecv _(cj) = 1500 km//h`
`therefore vecv _(c) = 1500 - 500 = 1000 km//h.`
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