Posted by ky on Thursday, July 26, 2012 at 5:26pm.
You can either use a formula or construct a table. You should get the same answer.
If the falling object drops 2.65 feet the first second, the value of the accleration of gravity there is
g' = 5.3 ft/s^2.
In t = 10 seconds, use the formula
Y = (g'/2)*t^2, with
g' = 5.3 ft/s^2 and t = 10 s.
Y = 265 feet.
1st second: 2.65 ft
2nd second: 5.3 + 2.65 = 7.95 ft
3rd second: 7.95 + 5.3 = 13.25 ft
4th second: 18.55 ft
5th second: 23.85 ft
6th second: 29.15 ft
7th second: 34.45 ft
8th second: 39.75 ft
9th second: 45.05 ft
10th second: 50.35 ft
Total: 262.35 ft
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