Question

A jet airplane travelling at the speed of $500km{h^{ - 1}}$ ejects its products of combustion at the speed of $1500km{h^{ - 1}}$ relative to the jet plane. What is the speed of the latter with respect to an observer on the ground?

Answer 1

Hint: There are two speeds which are given to us, the speed of the combustion of the particles and the speed of the jet airplane. The speed of the combustion particle is given with respect to the jet airplane and the speed of the particle is given with respect to the observer standing on the ground.

Complete step by step answer:
As we are given the speed of a jet airplane the speed of combustion article and we are asked to find the speed of the combustion particles with respect to the observer on the ground so here we need to understand each speed and with which respect it is given. The speed of the plane is with respect to the ground or we can say with respect to the observer on the ground, the speed of the combustion particle is given with respect to the plane and they are also in the opposite direction to each other. In order to calculate the speed of the combustion particle, we need to write the relation of the difference of the speeds of the combustion particle with respect to the observer on the ground and the speed of the jet airplane with respect to the observer on the ground. So the relation that we get is given by,
\[ \Rightarrow {V_{\dfrac{c}{p}}} = {V_c} - {V_p}\]
Where \[{V_{\dfrac{c}{p}}}\]is the speed of the combustion particle form the jet airplane, \[{V_c}\] is the speed of the combustion particle with respect to the observer on the ground and \[{V_p}\] is the speed of the jet airplane with respect to the observer on the ground.
$ \Rightarrow 1500 = {V_c} - \left( {500} \right)$
Solving for $V_c$
$ \Rightarrow 1500 = {V_c} - \left( { - 500} \right)$
$ \Rightarrow 1500 = {V_c} + 500\\$ On simplification,
$ \Rightarrow {V_c} = 1000\dfrac{{km}}{h}$

Therefore, the velocity of the combustion particle is equal to${V_c} = 1000\dfrac{{km}}{h}$.

Note:
As the hot combustion gases comes out in order to conserve the linear momentum the jet airplane starts moving equally the equal momentum as of the combustion particle but in the opposite direction.