Question

A jet airplane travelling at the speed of 500 km/h ejects its products of combustion at the speed of 1500 km/h relative to the jet plane. What is the speed of the latter with respect to an observer on the ground??​

Answer 1

Answer:

Relative velocity

V

A/B

=v A−v B

given

V

A/B=1500Km/h

v

A

−v

B=1500

v

A

−(−500)=1500

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

V

A

=1000km/h

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,

$\Large\vec{v}_{j}\:= \: 500\:Km\:h^{-1}$

$\Large\vec{v}_{cj}\:=\: -1500 \:Km\:h^{-1}h$

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

To Find :-

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

Solution:-

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

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.