Answer
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Hint: We have a ball that is dropped from a height. The ball reaches the ground and rebounds. We need to find the height to which the ball rebounds. We know the equation for velocity of a body falling from a height and a relation between velocity and coefficient of restitution. Substituting values in these equations we get the required solution.
Formula Used:
Velocity of a body falling from a height ‘H’ is given by,
$v=\sqrt{2gH}$
Coefficient of restitution and velocity,
${{v}_{f}}=ev$
Complete step-by-step answer:
In the question a ball is dropped from a height 1m
This is the initial height of the ball.
Let us denote this height by ‘H’
H = 1m
The ball reaches the ground and bounces from the surface.
Let the height to which the ball rebounds be denoted as ‘h’
We know that when a body falls from a height ‘H’, the velocity of that body is given by the equation,
$v=\sqrt{2gH}$
In this case, value of H is 1m
Therefore, velocity of the ball will be
$v=\sqrt{2g}$
This is the velocity of the ball when it strikes the ground.
After striking the ground the ball will rebound with a velocity. Let this velocity be the final velocity.
Final velocity is given by the equation,
${{v}_{f}}=ev$, were ‘${{v}_{f}}$’ is final velocity, ‘e’ is coefficient of restitution and ‘v’ is the striking velocity.
When the ball strikes the ground and rebounds to a height h, the velocity with which the body rebounds, i.e. the final velocity will become,
${{v}_{f}}=\sqrt{2gh}$
Since ${{v}_{f}}=ev$, we get
$\begin{align}
& \sqrt{2gh}=e\sqrt{2g} \\
& 2gh={{e}^{2}}2g \\
& h={{e}^{2}} \\
\end{align}$
In the question we are given the value of coefficient of restitution, e = 0.6
Therefore the height to which the ball rebounds ‘h’ will become,
$\begin{align}
& h={{0.6}^{2}} \\
& h=0.36m \\
\end{align}$
Hence the correct answer is option C.
Note: When two objects collide, the coefficient of restitution is the ratio of the final velocity to the initial relative velocity of the two objects after collision.
In the case of a bouncing ball, the coefficient of restitution is calculated by taking the square root of the ratio of one bounce to the height of the next bounce.
Formula Used:
Velocity of a body falling from a height ‘H’ is given by,
$v=\sqrt{2gH}$
Coefficient of restitution and velocity,
${{v}_{f}}=ev$
Complete step-by-step answer:
In the question a ball is dropped from a height 1m
This is the initial height of the ball.
Let us denote this height by ‘H’
H = 1m
The ball reaches the ground and bounces from the surface.
Let the height to which the ball rebounds be denoted as ‘h’
We know that when a body falls from a height ‘H’, the velocity of that body is given by the equation,
$v=\sqrt{2gH}$
In this case, value of H is 1m
Therefore, velocity of the ball will be
$v=\sqrt{2g}$
This is the velocity of the ball when it strikes the ground.
After striking the ground the ball will rebound with a velocity. Let this velocity be the final velocity.
Final velocity is given by the equation,
${{v}_{f}}=ev$, were ‘${{v}_{f}}$’ is final velocity, ‘e’ is coefficient of restitution and ‘v’ is the striking velocity.
When the ball strikes the ground and rebounds to a height h, the velocity with which the body rebounds, i.e. the final velocity will become,
${{v}_{f}}=\sqrt{2gh}$
Since ${{v}_{f}}=ev$, we get
$\begin{align}
& \sqrt{2gh}=e\sqrt{2g} \\
& 2gh={{e}^{2}}2g \\
& h={{e}^{2}} \\
\end{align}$
In the question we are given the value of coefficient of restitution, e = 0.6
Therefore the height to which the ball rebounds ‘h’ will become,
$\begin{align}
& h={{0.6}^{2}} \\
& h=0.36m \\
\end{align}$
Hence the correct answer is option C.
Note: When two objects collide, the coefficient of restitution is the ratio of the final velocity to the initial relative velocity of the two objects after collision.
In the case of a bouncing ball, the coefficient of restitution is calculated by taking the square root of the ratio of one bounce to the height of the next bounce.
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