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# A packet dropped from a balloon going upwards with a velocity of $12ms^{-1}$, the velocity of the packet after 2 seconds will be:$\text {A}.\ -12ms^{-1}$$\text {B}.\ 12ms^{-1}$$\text {C}.\ 7.6ms^{-1}$$\text {D}.\ -7.6ms^{-1}$

Last updated date: 13th Jun 2024
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Hint: Since the balloon must be flying in the atmosphere, hence it must be experiencing a constant gravitational force towards the earth. After the packet is dropped, the only force acting upon the packet is gravity (neglecting air friction). Hence the net acceleration of the packet is $g=9.8ms^{-2}$. Also since the acceleration is towards downwards, so we'll take it with a negative sign.

Formula used: $v=u+at$, where v is the final velocity of the particle having acceleration ‘a’ after time ‘t’, and ‘u’ is its initial velocity.

Given that the packet was also moving up along with the balloon. Hence the velocity of both the packet and balloon must be the same. Hence $u =12ms^{-1}$,
$g=-9.8ms^{-2}$ and time t=2sec.
Now using the equation of motion: $v=u+at$, we get
$v=12 + (-9.8)\times 2$
Or $v=12-19.6=-7.6\ ms^{-1}$
Hence the magnitude of the final velocity of the particle is$7.6\ ms^{-1}$. It is important to note that negative sign shows that the velocity is in the direction of ‘g’ or opposite to the direction of movement of the balloon i.e. it’s direction is downwards.
Note: One should note here that the reason of using the equation $v=u+at$, instead of $s=ut+\dfrac 12 at^2$ or $v^2-u^2=2as$ is that there is nowhere in the question we are dealing with distances or displacement. One can use them to relate velocity and displacement, when time is not given ($v^2-u^2=2as$) and when we are asked to find displacement in given time and final velocity is not given ($s=ut+\dfrac 12 at^2$).