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July 24, 2014

July 24, 2014

Posted by **ky** on Thursday, July 26, 2012 at 5:26pm.

- physics -
**drwls**, Thursday, July 26, 2012 at 6:06pmYou 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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