Question

If a ball is dropped from h height of 2.25 m on a smooth floor attains the height of the bounce equal to 1.00m the coefficient of the restitution between the ball and the floar is

Answer 1

........................

Answer 2

Given that,

Initial position of the ball, h = 2.25 m

Final position of the ball, H = 1 m

To find,

The coefficient of the restitution between the ball and the floor.

Solution,

The ratio of speed of separation to the speed of approach is called coefficient of restitution. The speed of the ball can be calculated using conservation of energy for two positions such that :

$u=\sqrt{2gh} =\sqrt{2\times 10\times 2.25} =6.7\ m/s$

and

$v=\sqrt{2gH} =\sqrt{2\times 10\times 1} =4.47\ m/s$

The coefficient of the restitution is given by :

$e=\dfrac{v}{u}\\\\e=\dfrac{4.47}{6.7}\\\\e=0.66$

So, the coefficient of the restitution between the ball and the floor is 0.66.

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Coefficient of the restitution

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