Question

A jet plane is moving vertically upward with speed 300 kilometre per hour it is rejecting product of combustion with speed 1200 kilometre per hour with respect to find out actual product of combustion.​

Answer 1

Answer:

900 km per hour

Explanation:

Given

Speed of moving jet plane = 300 kilometre per hour

Speed of ejection product combustion = 1200 kilometre per hour with respect to itself

A.T.Q.

=> Vr = Vj + VP

=> 1200 = 300 + Vp

=> Vp = 900 km/ h

Hence

Speed of combustion product is 900 km per hour.

Answer 2

=》Velocity of jet wrt latter=300 km/h

Velocity of jet wrt latter=300 km/hVelocity of latter wrt jet= -1200km/h  (-ve sign indicates the motion in opposite                                                                    direction)

Velocity of jet wrt latter=300 km/hVelocity of latter wrt jet= -1200km/h  (-ve sign indicates the motion in opposite                                                                    direction)Velocity of latter w.r.t groung = 300+(-1200)

Velocity of jet wrt latter=300 km/hVelocity of latter wrt jet= -1200km/h  (-ve sign indicates the motion in opposite                                                                    direction)Velocity of latter w.r.t groung = 300+(-1200)                                               =-900km/h

Velocity of jet wrt latter=300 km/hVelocity of latter wrt jet= -1200km/h  (-ve sign indicates the motion in opposite                                                                    direction)Velocity of latter w.r.t groung = 300+(-1200)                                               =-900km/h-ve sign shows opposite  direction  required  Velocity is 900 km per hour