假設我在100m的半空放下一舊石頭, 它到達地面需要多少時間??

s = 100m , u = 0ms-1 , a = 9.81ms-2
利用公式 s = u + (1/2)at2去計，
s = u + (1/2)at2
100 = 0 + (1/2)(9.81)(t)2
100 = (4.905)(t)2
100 / 4.905 = t2
t2 = 20.387
t = 4.52s
它到達地面需要4.52s

2008-04-23 19:57:41 補充：
為了方便計算，通常都會設a = g(重力加速度) = 10ms^-2
s = u + (1/2)at^2
100 = 0 + (1/2)(10)(t)^2
100 = (5)(t)^2
100 / 5 = t^2
t^2 = 20
t = 4.47s
它到達地面需要4.47秒

2008-04-23 20:00:18 補充：
t^2 = 20.387
t = ±√20.387
t = 4.52 或 -4.52(捨去，因所求的時間不會負數)
所以t = 4.52s