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