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. What is the speed of the later with respect to an observer on the ground?

Answer 1

Hey mate,

# Answer: vp = -1000 km/h

# Given-
- Speed of jet plane
vj = 500 km/h
- Speed of ejection of products
vp = ? km/h
- Relative speed of ejection of products wrt jet plane
vpj = -1500 km/h

# Solution-
Relative speed of ejection of combustion products wrt jet plane is given by formula,
vpj = vp-vj
-1500 = vp-500
vp = -1000km/h

Speed of ejection of combustion products observed from ground is 1000km/h in direction opposite to the direction of motion of the jet airplane.

Hope that ia helpful...

Answer 2

Let Vp be the velocity of products w.r.t ground

Let us consider the direction of jet plane to be positive  

Speed of jet plane =va= 500 km/h

Relative speed of products combustion w.r.t jet plane =VpA=1500km/h

Relative velocity of products w.r.t airplane = Vp- Va= -1500

vp=500-1500

= - 1000 km/h

Here negative direction shows that the direction of products of combustion is opposite to that of airplane.

∴ The magnitude of Relative velocity of products w.r.t airplane is 1000km/h