Question

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

• A. 500 km h^-1
• B. 1000 km h^-1
• C. 1500 km h^-1
• D. 2000 km h^-1

Answer 1

Answer: (B)

1000 km h^-1

Answer 2

Velocity of jet plane w.r.t. ground $V_{JG} = 500kmh^{- 1}$

• ve

Velocity of products of combustions w.r.t. jet plane $V_{CJ} = - 1500kmh^{- 1}$

$\therefore$Velocity of products of combustions w.r.t. ground is $V_{CG} = V_{CJ} + V_{JG} = 1500kmh^{- 1} + 500kmh^{- 1} = - 1000kmh^{- 1}$

- ve sing shows that the direction of products of combustion is opposite to that of the plane.

$\therefore$Speed of the products of combustions w.r.t. ground = $1000kmh^{- 1}.$