Question

a jet air plane traveling at a speed of 750 km/h ejects gases in opposite direction at a speed of 2000 km/h relative to jet plane.Calculate the velocity of gases with respect to earth.​

Answer 1

Answer:

Velocity of jet plane = 750 km/h

Velocity of  gas of jet plane = -2000km/h  (- ve sign indicates the motion in                                                                                       opposite    direction)

Velocity of latter w.r.t groung = 750 +(-2000)

                                              = -1250km/h

                                            ( -ve sign shows opposite  direction )

Hence  required  Velocity is 1250km/h

Hope it helps your needs ^_^ .......

Answer 2

Question:-

A jet airplane travelling at the speed of 400 km\h ejects gases at a speed of 800km\h the speed of the gas relative to rocket is or (speed of latter by seen by a observer in the ground.)

Explanation:

Given,

$\Large\vec{v}_{j}\:= \: 750\:Km\:h^{-1}$

$\Large\vec{v}_{cj}\:=\: -2000 \:Km\:h^{-1}h$

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

To Find :-

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

Solution:-

$\vec{v} _{cj} =\vec{v}_{c} - \vec{v}_{j} \\ - 2000 =\vec {v} _{c} - 750 \\ - 2000 + 750 = \vec{v} _{c} \\ ➜\vec{v} _{c} = - 1250$

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 1250 Km\h in opposite direction in plane.

The speed of gas is 1250km\hr or latter

Or,

The speed observe by the observer is 1250 Km\h.

Here, All statements are correct.