
From a balloon rising vertically upward at $5\,m/s$ a stone is thrown up at $10\,m/s$ relative to the balloon. Its velocity with respect to the ground after $2s$ is. $(g = 10\,m/{s^2})$
(A) $0\,m/s$
(B) $20\,m/s$
(C) $10\,m/s$
(D) $5\,m/s$
Answer
149.7k+ views
Hint: We will find the initial velocity of the balloon which is the addition of the initial velocity of the stone and the balloon as given in the question. Next, we will use the displacement equation for time $t = 2s$ to get the displacement of the stone. We will calculate the velocity using the displacement in the equation ${v^2} = {u^2} + 2aS$.
Formula used: Laws of motion equation
$S = ut + \dfrac{1}{2}a{t^2}$ and
${v^2} = {u^2} + 2aS$.
Complete step by step answer::
It is given that the balloon is rising vertically upwards with a velocity of $5\,m/s$. This is the initial velocity of the balloon and it does not change. From this balloon, we throw a stone at a speed of $10\,m/s$ upwards, in the same direction as that in which the balloon is travelling.
Thus the initial velocity of the stone is $15\,m/s$, since the speed of the balloon will get added to it. Now we have to find the velocity of the balloon at the time $t = 2s$. For this, we will use the laws of motion equation
$S = ut + \dfrac{1}{2}a{t^2}$,
where $S$ is the required displacement.
Substituting the value of $t = 2s$ and the initial velocity of $u = 15\,m/s$ in the above equation, we get $S = (15\,m/s)(2s) + \dfrac{1}{2}( - 10\,m/{s^2}) \times {(2s)^2}$
$ \Rightarrow S = 20\,m$.
Next, we will use
${v^2} = {u^2} + 2aS$, where
$ \Rightarrow {v^2} = {(15\,m/s)^2} + 2( - 10\,m/{s^2})(20\,m) = {(5\,m/s)^2}$
Thus the magnitude of the velocity is
$v = 5\,m/s$
This is the velocity of the stone with respect to the ground after $2s$.
Therefore, the correct answer is option (D).
Note: The initial velocity of the stone needs to be found carefully since the reference frame should be with respect to the ground. Thus we add the velocity of the balloon with respect to the ground with the velocity of the stone with respect to the balloon. Here we will add the two velocities and not subtract, since both the velocities are directed towards the same direction and not opposite or at any other angle.
Formula used: Laws of motion equation
$S = ut + \dfrac{1}{2}a{t^2}$ and
${v^2} = {u^2} + 2aS$.
Complete step by step answer::
It is given that the balloon is rising vertically upwards with a velocity of $5\,m/s$. This is the initial velocity of the balloon and it does not change. From this balloon, we throw a stone at a speed of $10\,m/s$ upwards, in the same direction as that in which the balloon is travelling.
Thus the initial velocity of the stone is $15\,m/s$, since the speed of the balloon will get added to it. Now we have to find the velocity of the balloon at the time $t = 2s$. For this, we will use the laws of motion equation
$S = ut + \dfrac{1}{2}a{t^2}$,
where $S$ is the required displacement.
Substituting the value of $t = 2s$ and the initial velocity of $u = 15\,m/s$ in the above equation, we get $S = (15\,m/s)(2s) + \dfrac{1}{2}( - 10\,m/{s^2}) \times {(2s)^2}$
$ \Rightarrow S = 20\,m$.
Next, we will use
${v^2} = {u^2} + 2aS$, where
$ \Rightarrow {v^2} = {(15\,m/s)^2} + 2( - 10\,m/{s^2})(20\,m) = {(5\,m/s)^2}$
Thus the magnitude of the velocity is
$v = 5\,m/s$
This is the velocity of the stone with respect to the ground after $2s$.
Therefore, the correct answer is option (D).
Note: The initial velocity of the stone needs to be found carefully since the reference frame should be with respect to the ground. Thus we add the velocity of the balloon with respect to the ground with the velocity of the stone with respect to the balloon. Here we will add the two velocities and not subtract, since both the velocities are directed towards the same direction and not opposite or at any other angle.
Recently Updated Pages
JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Uniform Acceleration

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Degree of Dissociation and Its Formula With Solved Example for JEE

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Motion in a Straight Line Class 11 Notes: CBSE Physics Chapter 2

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry
