Tennis balls are tested by being dropped from a height of 2.5 m onto a concrete

Question

Tennis balls are tested by being dropped from a height of 2.5 m onto a concrete floor. The 57 g ball hits the ground, compresses, then rebounds. A ball will be accepted for play if it rebounds to a height of about 1.4 m; it will be rejected if the bounce height is much more or much less than this.

Suppose a ball is dropped from 2.5 m and rebounds to 1.4 m. How fast is the ball moving just before it hits the floor?

What is the ball's speed just after leaving the floor?

What happens to the "lost" energy? The "lost" energy is transformed into?

If the time of the collision with the floor is 5.9 ms, what is the average force on the ball during the impact?

Explanation / Answer

speed before = 2gh = (2*9.8*2.5) = 7 m/s

speed after = 2gh = (2*9.8*1.4) = 5.2383 m/s

lost energy is transformed into majorly heat and may also be used in compressing the ball during rebounding

avg force = change in momentum/time = m(5.2383-(-7)) /0.0059 = 118.23 N

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