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I need help attacking this problem A steel ball is dropped onto a hard floor from a height of 1.95 m and rebounds to a height of 1.87 m (assume positive direction is upward)

ACaculate it's velocity just before it strikes the floor

Bcalculate it's velocity just after it leaves the floor on its way back up

CCalculate it's acceleration during contact with the floor if that contact lasts .0800ms

DHow much did the ball compress during its collision with the floor, assuming the floor is absolutely rigid

Verified Answer :

(1/2) m v^2 = m g h 
A.) v = sqrt (2 g h) = -sqrt (29.81.95) 
negative because down 

B.) (1/2)m v^2 = m g h 
or v = sqrt (29.81.87) 
positive because up) 

C.) change in velocity = (magnitude Part A + magnitude part Part B) 
divide that by .08 to get acceleration up 

D.) (1/2)(a from part C) (.08)^2

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