
A balloon is rising with a constant acceleration \[2m/{s^2}\]. At a certain instant when the balloon was moving with a velocity of $4m/s$, a stone was dropped from it in a region where $g = 10m/{s^2}$. The velocity and acceleration of stone as it comes out from the balloon are respectively (in m/s and $m/{s^2}$)
A. 4, 10
B. 4,5
C. 10, 4
D. None of these
Answer
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Hint: When the stone is inside the balloon, they both are in the same frame and have the same motion but when it is thrown out of the balloon, it is falling under influence of gravity alone and its acceleration does not depend on the acceleration of the balloon now.
Complete step-by-step answer:
We are given a balloon which is rising upwards with a constant acceleration of \[2m/{s^2}\]
A stone is contained in it while it is rising upwards. The stone is in the moving frame of reference of the balloon as long as it is inside the balloon. The objects in the same frame of reference move with the same velocity and acceleration. So, as it comes out of the balloon, it will have the same velocity as the balloon i.e. 4 m/s.
As the stone comes out of the balloon, it comes under the influence of gravity and starts falling. During this time, the acceleration of the stone will be the same as the acceleration due to gravity which is $10m/{s^2}$.
Hence the correct answer to the question would be option A as the velocity of the stone is 4m/s while its acceleration is equal to the acceleration due to gravity which is $10m/{s^2}$.
Note: The acceleration of the stone is equal to the upward acceleration of the balloon only as long as it is moving with the balloon. As soon as the stone leaves the frame of reference of the balloon it achieves the acceleration equal to the acceleration due to gravity.
Complete step-by-step answer:
We are given a balloon which is rising upwards with a constant acceleration of \[2m/{s^2}\]
A stone is contained in it while it is rising upwards. The stone is in the moving frame of reference of the balloon as long as it is inside the balloon. The objects in the same frame of reference move with the same velocity and acceleration. So, as it comes out of the balloon, it will have the same velocity as the balloon i.e. 4 m/s.
As the stone comes out of the balloon, it comes under the influence of gravity and starts falling. During this time, the acceleration of the stone will be the same as the acceleration due to gravity which is $10m/{s^2}$.
Hence the correct answer to the question would be option A as the velocity of the stone is 4m/s while its acceleration is equal to the acceleration due to gravity which is $10m/{s^2}$.
Note: The acceleration of the stone is equal to the upward acceleration of the balloon only as long as it is moving with the balloon. As soon as the stone leaves the frame of reference of the balloon it achieves the acceleration equal to the acceleration due to gravity.
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