A ball is dropped 2.75 m to the floor below. What is the acceleration of the ball while it is on the floor if it is on the floor for 0.0400 s?

Question

I'm really confused on this question because it is a free fall question so wouldn't the acceleration be -9.81 m/s/s? How is the answer +340.0 m/s/s?

This is what I have but I don't know which equation to use...
y = -2.75 m
vi = 0 m/s
vf = ? - m/s
a = -9.81 m/s/s
t = 0.0400 s

Answer 1

You have the right idea about getting the speed at which it hits the floor.
HOWEVER the velocity changes when it hits
acceleration = change of velocity / change in time
it is on the floor, so velocity goes from contact speed to zero in 0.04 seconds
now
contact speed:
a = -9.81 m/s^2
v = 0 - 9.81 t
0 = 2.75 + 0 - 4.9 t^2
so t = sqrt (2.75 / 4.9) = 0.749 seconds
so v = - 9.81* 0.749 = -7.35 m/sec t contact
NOW do the problem
a = [0 - ( -7.35) ] / time on floor (positive up)
Now of course I assumed the ball stopped. If it is a perfect ball and bounced UP at 7.35 m/s , then double the result because the change in velocity doubled

Answer 2

I tried it but I keep on getting the answer 367.5 m/s/s but the answer is +340.0 m/s/s what am I doing incorrectly?