Question

A balloon is rising with constant acceleration $2m{{s}^{-2}}$. Two stones are released from the balloon at the interval of 2sec. Find out the distance between the two stones 1sec after the release of the second stone
a) 48m
b) 84m
c) 40m
d) 60m

Answer 1

To find the distance between the two stones we will consider the frame of reference of the balloon. It is given to us in the question that the two stones are released at the interval of 2sec. After 1 sec of the release of the second stone, the first stone will have a total time of flight of 3 sec. For an observer let us say on the balloon, he will consider himself to be at rest. Hence we can consider the acceleration of the balloon upwards to be zero and downwards as its acceleration upwards. Hence using Newton’s second kinematic equation we will determine the distance between the two stones.

Formula used:
$S=Ut+\dfrac{1}{2}a{{t}^{2}}$

Complete answer:
It is given to us that the balloon is rising with constant acceleration $2m{{s}^{-2}}$. For an observer on the balloon he will consider himself to be at rest. Therefore we can say that his acceleration as well as his instantaneous velocity to be zero. When the stone is released, he will observe it as the stone moves down. Now according to our observer the stone will move down due to acceleration due to gravity (g)as well as he will say the ground is moving away by acceleration of $2m{{s}^{-2}}$. Hence the total acceleration of the stone downwards will be $(g+2m{{s}^{-2}})$. For a body under acceleration ‘a’, the displacement of the body in time ‘t’ with initial velocity ‘U’ is given by,
$S=Ut+\dfrac{1}{2}a{{t}^{2}}$
As per the frame of reference of the balloon, it is at rest. Therefore the initial velocity of the both the stones as they are released can also be considered to be at rest.
The time of flight of the first stone is 3 sec. Therefore the distance covered by the stone is,
$\begin{aligned} & {{S}_{1}}=Ut+\dfrac{1}{2}a{{t}^{2}} \\\ & \Rightarrow {{S}_{1}}=(0)3+\dfrac{1}{2}(10+2)\times {{3}^{2}} \\\ & \Rightarrow {{S}_{1}}=\dfrac{12\times 9}{2} \\\ & \Rightarrow {{S}_{1}}=54m \\\ \end{aligned}$
The time of flight of the second stone is 1 sec. Therefore the distance covered by the stone is,
$\begin{aligned} & {{S}_{2}}=Ut+\dfrac{1}{2}a{{t}^{2}} \\\ & \Rightarrow {{S}_{2}}=(0)1+\dfrac{1}{2}(10+2)\times {{1}^{2}} \\\ & \Rightarrow {{S}_{2}}=\dfrac{12}{2} \\\ & \Rightarrow {{S}_{2}}=6m \\\ \end{aligned}$
Hence the distance between the two stones is equal to,
$\begin{aligned} & S={{S}_{1}}-{{S}_{2}} \\\ & \Rightarrow S=54-6 \\\ & \Rightarrow S=48m \\\ \end{aligned}$

Therefore the correct answer of the above question is option a.

Note:
A simple way of understanding the frame of reference is as follows. The balloon is moving vertically upwards with an acceleration of $2m{{s}^{-2}}$. Add this acceleration of the balloon to the opposite direction. As a result the net acceleration of the balloon will be zero but for an observer on the balloon for him the ground will have an acceleration of $2m{{s}^{-2}}$ downwards. Therefore, for an observer on the balloon the stone will move downwards with acceleration due to gravity as well as the acceleration of the ground that is moving downwards.