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A rubber is dropped from a height (h=1.0m). After the first bounce (n=1), the ball reaches a lower maximum height of h1=0.8 m.

A) find the speed (v1) at which the ball bounced back up after hitting the floor.

B) find the percent of energy loss to heat, air resistance, and the deformation of the object.

C) predict the next maximum height (h2), if the “loss” of energy is constant.

D) how does v1 compared to v2

E) how does T1 compare to T2

A rubber is dropped from a height (h=1.0m). After the first bounce (n=1), the ball-example-1
User Paranoid
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A) To find the speed (v1) at which the ball bounced back up after hitting the floor, we can use the principle of conservation of energy. Since the ball reaches a lower maximum height, we know that some energy was lost. The speed at which it bounced back up can be found using the equation v1 = √(2gh1), where g is the acceleration due to gravity (approximately 9.8 m/s^2).

B) To find the percent of energy loss to heat, air resistance, and the deformation of the object, we can calculate the difference in potential energy before and after the bounce. The percent of energy loss can be determined by dividing the energy loss by the initial potential energy and multiplying by 100%.

C) If the "loss" of energy is constant, we can assume that the next maximum height (h2) will be lower than the previous maximum height (h1). However, without knowing the specific amount of energy loss, we cannot accurately predict the exact value of h2.

D) Without additional information or specific values for v2, we cannot compare v1 to v2.

E) Similarly, without specific values for T1 and T2, we cannot compare T1 to T2.
User Arcymag
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