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# 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

Last updated date: 17th Apr 2024
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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, $g = 10m/{s^2}$.