A jet airplane travelling at a speed of 500km/h ejects its products of combustion at the speed of 1500km/h relative to the jet plane. What is the speed of the latter with respect to an observer on the ground?
A
500km/h
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
1000km/h
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
1500km/h
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
2000km/h
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is B1000km/h
Let vjvg and v0 be the velocities of jet, ejected gases and observer on the ground respectively.
Let jet be moving towards right (+ve direction.) ∴ Ejected gases will move towards left (-ve direction). ∴ According to the statement vj−v0=500km/h……(i)
As observer is at rest vg−vj=−1500km/h(given)……(i) ∴ Adding Eqs. (i) and (ii), we get the speed of combustion products w.r.t., observer on the ground (vj−v0)+(vg−vj)=vg−v0=500+(−1500)
Or vg−v0=−1000km/h
-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.