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 ?​

Answer 1

Hey mate,

# Answer: vp = -1000 km/h

# Given-

- Speed of jet plane

vj = 500 km/h

- Speed of ejection of products

vp = ? km/h

- Relative speed of ejection of products wrt jet plane

vpj = -1500 km/h

# Solution-

Relative speed of ejection of combustion products wrt jet plane is given by formula,

vpj = vp-vj

-1500 = vp-500

vp = -1000km/h

Speed of ejection of combustion products observed from ground is 1000km/h in direction opposite to the direction of motion of the jet airplane.

Hope that ia helpful...

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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,

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

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

( 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.