Question

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

Answer 1

Find total steps forward and duration of time in one slot.
velocity of jet $={\nu }_A$
velocity of products of combustion $={\nu }_B$

Calculate the speed of the products of combustion.
Given speed of jet airplane, ${\nu }_A=500km/h$
velocity of $v_B-v_A$ will be in the negative direction.
Hence, $v_{BA}=v_B-v_A=-1500km/h$
Therefore, $-1500=v_B-500\Rightarrow v_B=-1000km/h$
The negative sign indicates that the direction of its products of combustion is opposite to the direction of the motion of the jet airplane.

Final Answer: The speed of the products of combustion is $1000km/h$ with respect to a stationary observer on the ground.