
A boy throws a stone from the top of a cliff into the sea. When the stone leaves his hand it is horizontally travelling at a velocity of $10m/s$. How long will it take the stone to reach the sea, a distance $8m$ below?
(A) $1\sec $
(B) $1.26\sec $
(C)$1.5\sec $
(D) None of these
Answer
126.3k+ views
Hint: The vertical and horizontal motions are independent of each other. The vertical displacement is given by Newton’s law of motion where acceleration is equal to the acceleration due to gravity. The initial velocity is zero for the stone in this case.
Formula Used: The formulae used in the solution are given here.
$S = ut + \dfrac{1}{2}a{t^2}$ where $S$ is the displacement, $u$ is the initial velocity, $t$ is the time taken for that displacement and $a$ is the acceleration.
Complete Step by Step Solution: It has been given that a boy throws a stone from the top of a cliff into the sea. When the stone leaves his hand it is horizontally travelling at a velocity of $10m/s$. The sea is at a distance $8m$ below.
Projectile motion is the motion of an object thrown or projected into the air, subject to only the acceleration of gravity. The object is called a projectile, and its path is called its trajectory. The motions along perpendicular axes are independent and thus can be analysed separately.
Then, resolve the position and/or velocity of the object in the horizontal and vertical components. The components of position s are given by the quantities $x$ and $y$, and the components of the velocity v are given by ${v_x} = v\cos \theta $ and ${v_y} = v\sin \theta $, where $v$ is the magnitude of the velocity and $\theta $ is its direction.
Now, the sea is vertically down the cliff. The vertical distance $y$ is given by the equation of motion for displacement.
By Newton’s law of motion, we have, $S = ut + \dfrac{1}{2}a{t^2}$ where $S$ is the displacement, $u$ is the initial velocity, $t$ is the time taken for that displacement and $a$ is the acceleration.
For vertical motion, the acceleration acting on the body is the acceleration due to gravity. It is represented by $g$.
Thus, $a = g$.
The displacement along the y-axis is given by, $S = y$.
Then, the formula for vertical displacement is given by, $y = ut + \dfrac{1}{2}g{t^2}$.
In this case, the initial velocity $u = 0$, since the stone is at rest before it is thrown.
Let the acceleration due to gravity is $g = 10m/{s^2}$.
The vertical displacement is given by, $y = 0 \times t + \dfrac{1}{2} \times 10 \times {t^2}$. The distance of the sea is $8m$. Thus, $y = 8m$.
Then, we can write,
$8 = 0 \times t + \dfrac{1}{2} \times 10 \times {t^2}$
$ \Rightarrow 8 = 5{t^2}$
The time taken for the stone to reach the sea is $t$ and it’s value is given by,
$t = \sqrt {\dfrac{8}{5}} $
$ \Rightarrow t = 1.26\sec $
Hence, the time taken is $1.26\sec $.
The correct answer is Option B.
Note: The vertical motion of a projectile is controlled by the force of gravity. This means that there is an unbalanced force acting on the ball and so the ball will accelerate downwards. This acceleration is $9.8m{s^{ - 2}} \simeq 10m{s^{ - 2}}$. (the gravitational field strength on Earth).
Note that the result of this is different if the value of acceleration due to gravity is not on the basis of the assumption that, $g = 10m/{s^2}$.
Formula Used: The formulae used in the solution are given here.
$S = ut + \dfrac{1}{2}a{t^2}$ where $S$ is the displacement, $u$ is the initial velocity, $t$ is the time taken for that displacement and $a$ is the acceleration.
Complete Step by Step Solution: It has been given that a boy throws a stone from the top of a cliff into the sea. When the stone leaves his hand it is horizontally travelling at a velocity of $10m/s$. The sea is at a distance $8m$ below.
Projectile motion is the motion of an object thrown or projected into the air, subject to only the acceleration of gravity. The object is called a projectile, and its path is called its trajectory. The motions along perpendicular axes are independent and thus can be analysed separately.
Then, resolve the position and/or velocity of the object in the horizontal and vertical components. The components of position s are given by the quantities $x$ and $y$, and the components of the velocity v are given by ${v_x} = v\cos \theta $ and ${v_y} = v\sin \theta $, where $v$ is the magnitude of the velocity and $\theta $ is its direction.
Now, the sea is vertically down the cliff. The vertical distance $y$ is given by the equation of motion for displacement.
By Newton’s law of motion, we have, $S = ut + \dfrac{1}{2}a{t^2}$ where $S$ is the displacement, $u$ is the initial velocity, $t$ is the time taken for that displacement and $a$ is the acceleration.
For vertical motion, the acceleration acting on the body is the acceleration due to gravity. It is represented by $g$.
Thus, $a = g$.
The displacement along the y-axis is given by, $S = y$.
Then, the formula for vertical displacement is given by, $y = ut + \dfrac{1}{2}g{t^2}$.
In this case, the initial velocity $u = 0$, since the stone is at rest before it is thrown.
Let the acceleration due to gravity is $g = 10m/{s^2}$.
The vertical displacement is given by, $y = 0 \times t + \dfrac{1}{2} \times 10 \times {t^2}$. The distance of the sea is $8m$. Thus, $y = 8m$.
Then, we can write,
$8 = 0 \times t + \dfrac{1}{2} \times 10 \times {t^2}$
$ \Rightarrow 8 = 5{t^2}$
The time taken for the stone to reach the sea is $t$ and it’s value is given by,
$t = \sqrt {\dfrac{8}{5}} $
$ \Rightarrow t = 1.26\sec $
Hence, the time taken is $1.26\sec $.
The correct answer is Option B.
Note: The vertical motion of a projectile is controlled by the force of gravity. This means that there is an unbalanced force acting on the ball and so the ball will accelerate downwards. This acceleration is $9.8m{s^{ - 2}} \simeq 10m{s^{ - 2}}$. (the gravitational field strength on Earth).
Note that the result of this is different if the value of acceleration due to gravity is not on the basis of the assumption that, $g = 10m/{s^2}$.
Recently Updated Pages
JEE Main 2023 (April 8th Shift 2) Physics Question Paper with Answer Key

JEE Main 2023 (January 30th Shift 2) Maths Question Paper with Answer Key

JEE Main 2022 (July 25th Shift 2) Physics Question Paper with Answer Key

Classification of Elements and Periodicity in Properties Chapter For JEE Main Chemistry

JEE Main 2023 (January 25th Shift 1) Maths Question Paper with Answer Key

JEE Main 2023 (January 24th Shift 2) Chemistry Question Paper with Answer Key

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

JEE Main Login 2045: Step-by-Step Instructions and Details

Class 11 JEE Main Physics Mock Test 2025

JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Keys & Solutions

JEE Mains 2025 Correction Window Date (Out) – Check Procedure and Fees Here!

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

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

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Fluids

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line
