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

Question

A jet airplane travelling at the speed of 500 km h -1 ejects its products of combustion at the speed of 1500 kmh -1 relative to the jet plane. The speed of the products of combustion with respect to an observer on the ground is (A) 500 kmh -1 (B) 1000 kmh -1 (C) 1500 kmh -1 (D) 2000 kmh -1

Tardigrade Admin · Accepted Answer

vC J=-1500 kmh -1=vC-vJ or vC=vC J+vJ=(-1500 kmh -1)+500 kmh -1 =-1000 kmh -1

Q. 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

$500 km h^{- 1}$

$1000 km h^{- 1}$

$1500 km h^{- 1}$

$2000 km h^{- 1}$

$v_{C J} = - 1500 km h^{- 1} = v_{C} - v_{J}$
or $v_{C} = v_{C J} + v_{J} = (- 1500 km h^{- 1}) + 500 km h^{- 1}$
$= - 1000 km h^{- 1}$

Q. A jet airplane travelling at the speed of 500kmh−1 ejects its products of combustion at the speed of 1500kmh−1 relative to the jet plane. The speed of the products of combustion with respect to an observer on the ground is

500kmh−1

1000kmh−1

1500kmh−1

2000kmh−1