Question

A ball is dropped from a height of 1m on a smooth floor. The height of first bounce is 0.810m. Determine, (a) coefficient of the restitution, (b) expected height of

second bounce.​

Answer 1

Answer:

The ans is in the attachment, plzz have a look.

Answer 2

(a) 0.9  (b) 0.6561 m

Given:

Height from which ball is dropped = 1m

The height of first bounce = 0.810m

To find:

(a) coefficient of the restitution

(b) expected height of second bounce.​

Solution:

When the ball is dropped from a height of 1 m, its velocity immediately before hitting the ground can be found by this equation:

$v^{2}$= $u^{2}$ + 2gh

Where,

u = 0 m / sec

g = 9.8 m / $sec^{2}$

h = 1 m

So, $v^{2}$= 0 + 2 x 9.8 x 1 = 19.6

=> v = (√19.6) m / sec

=> v ≈ 4.427 m / sec

Now, for the first bounce:

$v^{2}$= $u^{2}$ - 2gh

Where,

v = 0 m / sec

g = 9.8 m / $sec^{2}$

h = 0.81 m

So, 0 = $u^{2}$ - 2 x 9.8 x 0.81

=> u = {√(19.6 x 0.81)} m / sec

So, coefficient of restitution = {√(19.6 * 0.81)} / (√19.6) = 0.9

Since coefficient of restitution is 0.9

• Every next bounce will make the ball rise 0.9 times that of the previous bounce.

So, at first bounce the ball rises {$0.9^{2}$  x 1} m

= 0.81

So, at second bounce the ball should rise {$0.9^{2}$ x 0.81} m

= 0.6561 m.

Hence, coefficient of the restitution = 0.9 and expected height of

second bounce = 0.6561 m

Learn more about similar problems:

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