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Vedmedyk [2.9K]
2 years ago
10

A tennis ball is dropped from 1.3 m above

Physics
1 answer:
Dominik [7]2 years ago
4 0

Answer:

2.65m/s

Explanation:

Using the equation of motion:

v² = u²+2a∆S where

v is the final velocity

u is the initial velocity

∆S is the change in distance

a is the acceleration

Given

u = 0m/s

a = 9.8m/s²

∆S = 1.3-0.943

∆S = 0.357m

Substituting the given parameters into the formula

v² = 0²+2(9.8)(0.357)

v² = 0+6.9972

v² = 6.9972

v=√6.9972

v = 2.65m/s

Hence the velocity at which it hit the ground is 2.65m/s

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A tennis ball is released from a height of 4.0 m above the floor. After its third bounce off the floor, it reaches a height of 1
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Answer:

The percentage of its mechanical energy does the ball lose with each bounce is 23 %

Explanation:

Given data,

The tennis ball is released from the height, h = 4 m

After the third bounce it reaches height, h' = 183 cm

                                                                       = 1.83 m

The total mechanical energy of the ball is equal to its maximum P.E

                                      E = mgh

                                          = 4 mg

At height h', the P.E becomes

                                      E' = mgh'

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The percentage of change in energy the ball retains to its original energy,

                                 \Delta E\%=\frac{1.83mg}{4mg}\times100\%

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The ball retains only the 45% of its original energy after 3 bounces.

Therefore, the energy retains in each bounce is

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The ball retains only the 77% of its original energy.

The energy lost to the floor is,

                                E = 100 - 77

                                   = 23 %

Hence, the percentage of its mechanical energy does the ball lose with each bounce is 23 %      

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3 years ago
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