question_answer

A jet airplane travelling from east to west at a speed of 500\[km\,{{h}^{-1}}\]ejected out gases of combustion at a speed of 1500\[km\,{{h}^{-1}}\] with respect to the jet plane. What is the velocity of the gases with respect to an observer on the ground?

A)  1000 \[km\,{{h}^{-1}}\] in the direction west to east

B)  1000 \[km\,{{h}^{-1}}\] in the direction east to west

C)  2000 \[km\,{{h}^{-1}}\] in the direction west to east

D)  2000 \[km\,{{h}^{-1}}\] in the direction east to west

Correct Answer: A

Solution :

[a] \[\vec{v}A=-500\hat{i}\]and \[\vec{v}GA=1500\hat{i}\] \[\Rightarrow \vec{v}GA+\vec{v}A=\vec{v}GAS\] \[\Rightarrow \vec{v}GAS=1500\hat{i}-500\hat{i}=1000\hat{i}\]