Question

A jet airplane travelling at the speed of 450 km/h ejects the burnt gases at the speed of 1200 km/h relative to the jet airplane. find the speed of burnt gases​

Answer 1

Answer:

may this help you

Explanation:

Velocity of jet w.r.t. ground Vjg= 450 km/h …...(Upward)

Velocity of products relative to jet VPJ = 1200 km/h …(downward)

Hence, velocity of products relative to ground = -1200+450

= -750 km/h

Here – ve sing means downward .

Answer 2

Question:-

A jet airplane travelling at the speed of 500 km\h ejects gases at a speed of 1000km\h the speed of the gas relative to rocket is or (speed of latter by seen by a observer in the ground.)

Explanation:

Given,

～～～$\LARGE\vec{v}_{j}\:= \: 500\:Km\:\h$

～～～$\LARGE\vec{v}_{cj}\:=\: -1500 \:Km\:\h$

( Note:- here velocity of cumbustion or gas in negative direction)

To Find :-

$\vec{v}_{c}\:=\:?$

Solution:-

$\vec{v} _{cj} =\vec{v}_{c} - \vec{v}_{j} \\ - 1500 =\vec {v} _{c} - 500 \\ - 1500 + 500 = \vec{v} _{c} \\ ➜\vec{v} _{c} = - 1000$

Note:-

$\vec{v}_{cj}$ is Velocity of cumbustion with respect to jet.

$\vec{v}_{c}$ is Velocity of cumbustion respect to the ground which we are finding.

$\vec{v}_{j}$ is Velocity of jet.

Velocity of cumbustion in 1000 Km\h in opposite direction in plane.

The speed of gas is 1000km\hr or latter

Or,

The speed observe by the observer is 1000 Km\h.

Here, All statements are correct.