gravity related question this is. please know me about gravity force.
2007-02-03 06:19:36 · 4 answers · asked by jigneshpatel234 1 in Science & Mathematics ➔ Physics
(assuming there is no friction)
v1 = 0
t= ?
a = -9.8 m/s^2 (acceleration due to gravity)
di = 0 km
df= 1km
df = ½ at² + vt + di
0 m = 1/2 (-9.8 * t² ) + (0) t + 1000m
-1000 m = 1/2 (-9.8 * t² )
-2000 m = -9.8 * t²
204 = t²
14s= t
2007-02-03 06:35:09 · answer #1 · answered by pimpster 2 · 0⤊ 0⤋
The gravitational field is numerically equal to the acceleration of objects under its influence, and its value at the Earth's surface, denoted g, is approximately 9.8 m/s². This means that, ignoring air resistance, an object falling freely near the earth's surface increases in speed by 9.807 m/s (around 32 ft/s or 22 mph) for each second of its descent. Thus, an object starting from rest will attain a speed of 9.807 m/s (32.17 ft/s) after one second, 19.614 m/s (64.34 ft/s) after two seconds.
Now please you calculate the rest as I have no time.
2007-02-03 06:57:34 · answer #2 · answered by dragon77 2 · 0⤊ 0⤋
substitute your values in this formula
s=ut+1/2at^2
where 's' is the distance
'u' is the initial velocity i.e. 0
't' is the time
'a' is the accelaration due to gravity i.e. 9.8m/s^2
you will get an equation
1000m = 0*t+1/2*9.8*t^2
solve it, you will get the answer.
2007-02-03 17:07:56 · answer #3 · answered by Ray 2 · 0⤊ 0⤋
Where are your keepers?
2007-02-03 06:23:23 · answer #4 · answered by Vanimal 1 · 0⤊ 0⤋