2007-03-28 12:00:04 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Physics
Gravity on earth = 9.8 meters per second squared.
2007-03-28 12:07:43 · answer #1 · answered by robertspraguejr 4 · 0⤊ 0⤋
You have to be more specific. Do you want the formula for gravitational force? F = G x M1 x M2/ r^2 where G is the gravitational constant, M1 and M2 are the masses of the two objects and r is the distance between them. Or do you want how to calculate the speed of an object falling? V = 1/2 g x t^2 where g is 32 ft/s^2 (9.8 m/s^2) and t is the time in free fall.
2007-03-28 19:11:18 · answer #2 · answered by misoma5 7 · 0⤊ 0⤋
Acceleration due to gravity on the surface of the Earth is as follows:
In metric (SI) units:
g = -9.81 m / s^2
In Imperial (British) units:
g = -32.2 ft / s^2
They mean that an object in free-fall in a gravitational field of 1g will accelerate downwards (hence the minus signs) at a rate of 9.81 m/s or 32.2 ft/s, every second. Thus, if you dropped an object which was dense enough that air resistance was negligible, it would be traveling downward at a rate of 19.62 metres per second, or 70.63 km/h, two seconds after you released it.
2007-03-28 19:08:23 · answer #3 · answered by poorcocoboiboi 6 · 0⤊ 0⤋
9.8 feet per second squared.
Or nine point eight feet per second per second.
Each second that the object is falling , it falls another 9.8 feet per second faster.
2007-03-28 19:09:33 · answer #4 · answered by AviationMetalSmith 5 · 0⤊ 0⤋
9.8 m/s*s that's 9.8 meters per second with the seconds time unit squared.
2007-03-28 19:08:00 · answer #5 · answered by Anonymous · 0⤊ 0⤋
9.81m/sec/sec
2007-03-28 19:07:38 · answer #6 · answered by Anonymous · 1⤊ 0⤋