How far does a freely falling body travel during the first second of
its fall?
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d=at^2
d=9.8m/s X 1^2
d=9.8m
after 1 second it is going 9.8m/s, so to find the distance multiply the average speed by the time so (9.8 + 0 (the starting speed))/2 = 4.9m/s, the multiplied by 1s = 4.9m
s=(1/2)at^2 = 4.9m
At the Earth’s surface it falls 16 feet.