A tennis ball bouncing on a hard surface compresses and then rebounds. The details of the rebound are specified in tennis regulations. Tennis balls, to be acceptable for tournament play, must have a mass of 57.5 g. When dropped from a height of 2.5 m onto a concrete surface, a ball must rebound to a height of 1.4 m. During impact, the ball compresses by approximately 6 mm.
How fast is the ball moving when it hits the concrete surface? (Ignore air resistance.)

The motion of the ball is a free fall motion, so it means that the ball falls down under the effect of the force of gravity only. Therefore, it has a constant acceleration (acceleration of gravity, g), and we can use the following suvat equation:

[tex]v^2-u^2=2as[/tex]

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the displacement

For the ball in this problem, we have:

u = 0 (initial velocity, the ball is dropped from rest)

[tex]a=g=9.8 m/s^2[/tex] (acceleration of gravity)

s = 2.5 m (vertical displacement)

Solving for v, we find the velocity at which the ball hits the concrete surface:

[tex]v=\sqrt{u^2+2as}=\sqrt{0+2(9.8)(2.5)}=7 m/s[/tex]

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