Question

6. A jet airplane travelling at the speed of 400 km h ejects its products of combustion at the speed of
1200 km h relative to the jet plane. What is the speed of the latter with respect to an observer on the
ground?

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Answer 1

Hey Friend Ans:-800km/h

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Answer 2

Given,

The velocity of the jet plane, $v_j$ = 400 km/h

The velocity of the products of combustion w.r.t the jet, $v_{cj}$ = 1200 km/h

To find,

The velocity of the products w.r.t an observer on the ground, $v_g$

Solution,

velocity of jet = +400 km/h

velocity of products = -1200 km/h

now, the relation between the velocity of products w.r.t the jet, the velocity of the jet, and the velocity observed by the observer can be written as

$v_{cj}$ = $v_g$ - $v_j$

⇒$v_g$ = $v_{cj}$ + $v_j$

⇒$v_g$ = -1200 + 400

⇒$v_g$ = -800 km/h

∴ The velocity of the products of combustion w.r.t to an observer on the ground is -800 km/h.