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 with detailed explanation and diagram..​

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

Concept:

An object's relative velocity is its speed as determined by a different observer. It is the rate at which the position of one object in relation to another changes over time.

Given:

Speed of jet airplane = 400 km/hr

Eject products at speed = 1200 km/hr

Find:

We need to determine the speed of the products of combustions with respect to an observer on the ground.

Solution:

1. An object's relative velocity is its speed as determined by a different observer. It is the rate at which, over time, the relative locations of two objects change with respect to one another.

2. Relative velocity is given as, V = Va - Vb where V is the relative velocity of combustion products w.r.t jet plane, Va is the relative velocity of combustion products w.r.t. observer on the ground and Vb is the speed of the jet plane.

We have, V = 1200 km/hr

Therefore, Va - Vb = 1200 km/hr

We have, Vb as - 400

Therefore, 1200 = Va - (- 400)

Va = 1200 - 400

Va = 800 km/hr

Thus, the speed of the products of combustions with respect to an observer on the ground is 800 km/hr.

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