Question

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

• A. $500\text{km/h}$
• B. $1000\text{km/h}$
• C. $1500\text{km/h}$
• D. $2000\text{km/h}$

Answer 1

The correct option is B $1000\text{km/h}$


Let $v_j,v_g$ and $v_0$ be the velocities of jet, ejected gases and observer on the ground respectively.
Let jet be moving towards right (+ve direction.)
$\therefore$ Ejected gases will move towards left (-ve direction).
According to question
$v_j-v_0=500\text{km/h}$ ... (i)
As observer is at rest
$v_g-v_j=-1500\text{km/h}$ (given) ... (ii)
Adding Equ. (i) and (ii), we get the speed of combustion products w.r.t., observer on the ground
$(v_j-v_0)+(v_g-v_j)=v_g-v_0$
$=500+(-1500)$
or $v_g-v_0=-1000{\text{kmh}}^{-1}$
-ve sign shows that relative velocity of the ejected gases w.r.t. observer is towards left or in a direction opposite to the motion of the jet plane.