
A packet dropped from a balloon going upwards with a velocity of $12ms^{-1}$, the velocity of the packet after 2 seconds will be:
$\text {A}.\ -12ms^{-1}$
$\text {B}.\ 12ms^{-1}$
$\text {C}.\ 7.6ms^{-1}$
$\text {D}.\ -7.6ms^{-1}$
Answer
484.2k+ views
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.
Complete step by step answer:
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.
So, the correct answer is “Option D”.
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 ($s=ut+\dfrac 12 at^2$).
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.
Complete step by step answer:
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.
So, the correct answer is “Option D”.
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 ($s=ut+\dfrac 12 at^2$).
Recently Updated Pages
Master Class 11 Accountancy: Engaging Questions & Answers for Success

Glucose when reduced with HI and red Phosphorus gives class 11 chemistry CBSE

The highest possible oxidation states of Uranium and class 11 chemistry CBSE

Find the value of x if the mode of the following data class 11 maths CBSE

Which of the following can be used in the Friedel Crafts class 11 chemistry CBSE

A sphere of mass 40 kg is attracted by a second sphere class 11 physics CBSE

Trending doubts
10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

State and prove Bernoullis theorem class 11 physics CBSE

What organs are located on the left side of your body class 11 biology CBSE

The combining capacity of an element is known as i class 11 chemistry CBSE
