A ball falls freely from rest at a height of 2 meters. Observations indicate that the ball reaches the ground in .64 seconds.

Based on this information what is its acceleration?

vAve='ds/'dt=2/.64=3.125m/s

v0=0m/s

vf=2*vAve=6.25m/s

'dv=(vf-v0)=6.25m/s

aAve='dv/'dt

aAve=6.25/.64=9.8m/s^2=gravity

Is this consistent with an observation which concludes that a ball dropped from a height of 5 meters reaches the ground in 1.05 seconds?

'dt=1.05s

'ds=5m

vAve=5m/1.05s=5.25m/s

v0=0m/s initial

vf=10.5m/s final

'dv=vf-v0=10.5m/s

aAve='dv/'dt=10.5m/s /1.05s=10m/s^2 the two are not the same but within the allowance of +-.02 of uncertainty

Are these observations consistent with the accepted value of the acceleration of gravity, which is 9.8 m / s^2?

Yes they both are within the allowance of +-.02m/s^2 of uncertainty

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