A tennis ball is dropped from 1.21 m above the ground. It rebounds to a height of 0.82 m. With what velocity does it hit the ground? The acceleration of gravity is 9.8 m/s?

Question

Answer in m/s

Answer 1

I assume you mean what velocity it hits the ground the second time. Doing an energy balance, you obtain

E_{P, before} ^{= E}_{K, after} because there is no non-conservative work done, no friction, and kinetic energy before is 0 and potential energy after (with the ball at ground level) is 0. This leads to

mgh_b = 1/2mv_a² and v_a = sqrt(2gh_b) (which is true regardless of mass and path when there is no friction.)

Plug in the height of .82 m and obtain v_a in m/s (If the question is the velocity when it is hitting the ground the first time, use 1.21 m)

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