
From the top of a 64 metres high tower a stone is thrown upwards vertically with a velocity of 48 m/s. The greatest height (in metres) attained by the stone assuming the value of gravitational acceleration $g=32\text{ }m/{{s}^{2}}$ is,
(a)100
(b)88
(c)128
(d)112
Answer
626.7k+ views
Hint: Here, we are applying the formula ${{v}^{2}}-{{u}^{2}}=2as$ by keeping in mind that at maximum height velocity, $v=0.$ Then, find the height of the tower.
Complete Step-by-Step solution:
Let $u$ be the initial velocity, $v$ be the final velocity, $a$ be the acceleration due to gravity, s be the displacement, h be the greatest height attained by the stone.
Here, we are given that the height of the tower is 64m, Initial velocity $u=48$m/s, gravitational acceleration $g=32\text{ }m/{{s}^{2}}$.
Here, we have to apply the formula for equation of motion given by,
${{v}^{2}}-{{u}^{2}}=2as$ where $s$ is the displacement.
First, we have to calculate the value for $s$. From the above equation we will get:
$s=\dfrac{{{v}^{2}}-{{u}^{2}}}{2a}\text{ }....\text{ (1)}$
We know that after a maximum height the velocity starts reducing. So we can say that at a maximum height the velocity $v=0.$
Here, we are throwing the stone against the gravitational force, therefore we can say that acceleration will be negative, $a=-32\text{ }m/{{s}^{2}}$.
Therefore, our equation (1) becomes:
$\begin{align}
& s=\dfrac{{{0}^{2}}-{{48}^{2}}}{2\times -32} \\
& s=\dfrac{0-2304}{-64} \\
& s=\dfrac{-2304}{-64} \\
& s=36 \\
\end{align}$
Hence, we got $s=36\text{ }m$, but this is the displacement. We have to calculate the greatest height attained by the stone, which is obtained by adding the displacement with the height of the tower.
Therefore, we will get:
$\begin{align}
& h=36+64 \\
& h=100 \\
\end{align}$
Hence, we can say that the greatest height attained by the stone is $100\text{ }m$.
Therefore, the correct answer for this question is option (a).
Note: Here,we have to know that after a maximum height the velocity starts reducing, therefore velocity at maximum height is zero. There is also a possibility that we will take acceleration positive which may lead to incorrect answers since, since we are throwing stones against the gravitational force, the acceleration will be considered negative.
Complete Step-by-Step solution:
Let $u$ be the initial velocity, $v$ be the final velocity, $a$ be the acceleration due to gravity, s be the displacement, h be the greatest height attained by the stone.
Here, we are given that the height of the tower is 64m, Initial velocity $u=48$m/s, gravitational acceleration $g=32\text{ }m/{{s}^{2}}$.
Here, we have to apply the formula for equation of motion given by,
${{v}^{2}}-{{u}^{2}}=2as$ where $s$ is the displacement.
First, we have to calculate the value for $s$. From the above equation we will get:
$s=\dfrac{{{v}^{2}}-{{u}^{2}}}{2a}\text{ }....\text{ (1)}$
We know that after a maximum height the velocity starts reducing. So we can say that at a maximum height the velocity $v=0.$
Here, we are throwing the stone against the gravitational force, therefore we can say that acceleration will be negative, $a=-32\text{ }m/{{s}^{2}}$.
Therefore, our equation (1) becomes:
$\begin{align}
& s=\dfrac{{{0}^{2}}-{{48}^{2}}}{2\times -32} \\
& s=\dfrac{0-2304}{-64} \\
& s=\dfrac{-2304}{-64} \\
& s=36 \\
\end{align}$
Hence, we got $s=36\text{ }m$, but this is the displacement. We have to calculate the greatest height attained by the stone, which is obtained by adding the displacement with the height of the tower.
Therefore, we will get:
$\begin{align}
& h=36+64 \\
& h=100 \\
\end{align}$
Hence, we can say that the greatest height attained by the stone is $100\text{ }m$.
Therefore, the correct answer for this question is option (a).
Note: Here,we have to know that after a maximum height the velocity starts reducing, therefore velocity at maximum height is zero. There is also a possibility that we will take acceleration positive which may lead to incorrect answers since, since we are throwing stones against the gravitational force, the acceleration will be considered negative.
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