A jet airplance travelling at the speed of ` 500 km ^(-1)` ejects its products of combustion at the speed of ` 1500 km h^(-1)` relative to the jet plane. What is the speed of the burnt gases with respect to observer on the ground ?

Options

A. `500 km h^(-1)`
B. `1000 km h^(-1)`
C. `1500 km h^(-1)`
D. `2000 km h^(-1)`

Answer

B

Explanation

Velocity of jet plane w.r.t. ground
`v_(JG) = 500 km h^(-1)`
Velocity of products of combustion w.r.t. jet plane
`v_(GJ) = - 1500 km h^(-1)`
`therefore` Velocity of products of combustion w.r.t ground
`v_(CG) = v_(CJ) + v_(JG) = -1500 km h^(-1) + 500 km h^(-1) = -1000 km h^(-1)`
-ve sighn shows that the direction of products of combustion is opposite to that of the plane.
`therefore` Speed of the products of combustion w.r.t. ground = `1000 km h^(-1)` .