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A tennis ball isdropped from 1.52m abovethe ground. It rebounds to a height of 0.918 m. With what velocity does it hit the ground? The acceleration of gravity is 9.8 m/s2 . (Let down be negative.) Answer in units of m/s.

(part 2 of 3) With what velocity does it leave the ground? Answer in units of m/s.
(part 3 of 3) If the tennis ball were in contact with the ground for 0.0107 s, find the acceleration given to the tennis ball by the ground. Answer in units of m/s2.

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1 answer

  1. (1/2) m v^2 = m g h
    (1/2) v^2 = gh
    or
    v = sqrt (2 g h) (remember that for dropped object)
    so
    v at ground = -sqrt (2*9.8*1.52)

    now
    goes up .918 meter
    same deal in reverse
    v up at ground = +sqrt (2*9.8*.918)

    a = change in velocity/change in time
    = [ +sqrt (2*9.8*.918)- - sqrt (2*9.8*1,52) ] / 0.0107

    note - - is +

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