Question

A jet airplane travelling at the speed of 400 km h-1 ejects its products of combustion at
the speed of 1200 km h-1 relative to the jet plane. What is the speed of the latter with
respect to an observer on the ground ? plz ans correctly ​

Answer 1

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,

$\Large\vec{v}_{j}\:= \: 400\:Km\:h^{-1}$

$\Large\vec{v}_{cj}\:=\: -1200 \:Km\:h^{-1}h$

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

To Find :-

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

Solution:-

$\vec{v} _{cj} =\vec{v}_{c} - \vec{v}_{j} \\ - 1200 =\vec {v} _{c} - 400 \\ - 1200 + 400 = \vec{v} _{c} \\ ➜\vec{v} _{c} = - 800$

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 800 Km\h in opposite direction in plane.

The speed of gas is 800km\hr or latter

Or,

The speed observe by the observer is 800 Km\h.

Here, All statements are correct.

Answer 2

Answer:

The speed of the latter with

respect to an observer on the ground is 800 km/hr.

Explanation:

Given, the jet airplane travelling at the speed of 400 km/hr.

Its products of combustion at the speed of 1200 km h-1.

We know relative velocity$V=V_A-V_B....(1)$

Where,

$V_B =$Velocity of the jet.

$V_A =$Velocity of combustion respect to the ground which we are finding.

Now according to question,

$V=1200 \: km \: {h}^{ - 1}$

and$V_A = 400 \: km \: {h}^{ - 1}$

So from equation (1)

$1200=V_A-( - 400)$

Since the velocity of combustion products and plane are in opposite directions.

$V_A = 800 \: km \: {h}^{ - 1}$