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A jet aeroplane travelling with speed of $500km/hr$ ejects its products of combustion at the speed of $1500km/hr$ relative to the jet plane. What is the speed of the latter with respect to an observer on the ground ?

Last updated date: 20th Jun 2024
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Hint:-First we will find the velocity of aeroplane and it’s ejected product with respect to Earth. Then we will try to find the velocity of the product with respect to the plane by thinking if we are an observer in the aeroplane then what the velocity of the product will appear to us.

Complete Step by step explanation:
Velocity of aeroplane with respect to Earth is ${V_a} = 500km/hr$
Velocity of product with respect to Earth is ${V_b} = 1500km/hr$

Above diagram is with respect to Earth
Now, suppose you are in a given aeroplane, and you are observing the product falling. Since, you are also moving in the direction of the product, then the velocity of the product appearing to us should be lower than actual speed with respect to Earth.
With respect to body A, this diagram will be like,

Hence, the velocity of product with respect to plane will be ${V_p} = {V_b} - {V_a}$
On putting values, we get
${V_p} = 1500 - 500$
On solving we get,
${V_p} = 1000km/hr$

So, the velocity of the product with respect to the plane will be $1000km/hr$.

Note:Since, our velocity came to be positive, it means if we are in aeroplane, then the product ejected will seem to us, moving away from us in the same direction in which the plane is also moving. If it would be negative, then the product would be moving in the opposite direction of the plane with respect to the plane.