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Free fall in Earth's gravity?
How many feet will an object fall at Earth's surface, in ten seconds, disregarding air friction, assuming that g eguals one, at the beginning of the fall?
3 Answers
- 1 decade agoFavorite Answer
g=9.81m/s
the first second
1s=9.8 m/s=v 9.81m=td
2s=19.62 m/s=v 29.43m=td
3s=28.43 m/s=v 57.86m=td
4s=38.24 m/s=v 96.1m=td
5s=48.05 m/s=v 144.15m=td
6s=57.86 m/s=v 202.01m=td
7s=67.67 m/s=v 269.68m=td
8s=77.48 m/s=v 347.16m=td
9s=87.29 m/s=v 434.45m=td
10s=97.1 m/s=v 531.55m=td
td=total distance
s=second
v=velocity
- SteveLv 71 decade ago
On the other hand, if you really DO mean g is 1 (ft/sec²), the answer is
D = ½gt² = ½*1*10² = 50 ft
Maybe you should rename the planet.......
Someone above me is mixing English and metric units; a fatal error.
- 1 decade ago
I think you mean initial velocity is 1m/s? g is constant
s = ut + 0.5at^2
s = (1)(10) + 0.5(9.8)(10^2)
s = 10 + 0.5(9800)
s = approx. 500 metres
Thats about 1650 feet
