Question

A jet airplane is travelling at a speed of 1000km/h.it ejects its products at the speed of 800km/h relative to the jet plane.what is the speed of the latter with respect to an observer at the ground?

Answer 1

$V_{jg}$ = velocity of jet relative to ground observer = 1000 km/h

$V_{pj}$ = velocity of products relative to jet =- 800 km/h   (negative sign indicates that the products are ejected in opposite direction of the motion of jet)

$V_{pg}$ = velocity of products relative to ground

Using the equation

$V_{pg}$ = $V_{pj}$ - $V_{jg}$

$V_{pg}$ = - 800 - 1000

$V_{pg}$ = - 1800 km/h

the negative sign indicates the downward direction.

Since we have been asked for the speed , we ignore the negative sign.

so speed = 1800 km/h

Answer 2

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Speed of the jet airplane, v jet = 500 km/h

Relative speed of its products of combustion with respect to the plane,

v smoke = – 1500 km/h

Speed of its products of combustion with respect to the ground = v′smoke

Relative speed of its products of combustion with respect to the airplane,

vsmoke = v′smoke – vjet

1500 = v′smoke – 500

v′smoke = – 1000 km/h

The negative sign indicates that the direction of its products of combustion is opposite to the direction of motion of the jet airplane.

I hope, this will help you

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