要計算物件下降的所需時間是怎樣計算的?

有公式,中四五有講架

s = ut + (1/2) at^2
s = 下降距離, u = 初速(靜止放定係拋佢落黎)
a = 地心加速度( 9.8 ms^(-2) )
t = 時間

設由高處垂直向下以 10 m/s 速度拋 ( 成 36 公里每小時, 行車都係 50 度)
s = 200 m (高樓大廈)

200 = 10t + (1/2)(9.8)t^2
49t^2 + 100t - 2000 = 0
t = 5.45 s (好快到地)

如果靜止拋
200 = (1/2)(9.8)t^2
t = 6.39 s (咁快就可以跌死你)

要視乎有多少個力作用在該物件上, 比如物件上有一個降落傘從而減慢了下降的速度, 掉下來的瞬間是由靜止放下來的還是大大力的拋下來, 是不是先向上拋再掉下來?

不如假設你是由靜止放手然後物件掉下來, 以下計算不包括氣流、空氣阻力等問題, 只是用最簡單的方法計算。

s 是位移, 你可以當是物體下降的距離
u 是初始速度, 如以我們的假止由靜止放下來, u=0
a 是加速度, 這視乎有多少個力作用在物體上, 如以我們的假設只有物體本身的重量, a=10
t 是下降所需的時間
s = ut + (1/2)a*t*t
s = 0 + (1/2)(10)t*t
s = 5*t*t
t = sqrt(s/5)

物件下降所需的時間是 下降的距離, 除以5, 再開方。
這裡用的全都是S.I. unit, 即時間的單位是秒, 距離的單位是米。

I don't know if you refer to free fall. If that is the case, you may use the simple equation in kinetics, i.e.
s = ut + (1/2).at^2
where s is the downward distance travelled, u is the initital speed, a is the acceleration due to gravity (g=9.81 m/s2) and t is the time taken
If the object starts from rest, then u=0 m/2,
s = (1/2)(g)(t^2)
t = sqrt[2s/g], where sqrt stands for [square-root]
The above equation has the assumption that air resistance is neglected. If air resistance is taken into account, and which is usually assumed to be proportional to velocity. One needs to solve the following differential equation,
m[dv/dt] + kv= mg
where m is the mass of the falling object and k is a constant
The solution is
v = (v0)[1-exp(-kt/m)]
where (v0) is the terminal velocity given by mg/k
The distance travelled s is found by integrating the expression for v with respect to time t.
But in this case, the time taken to travel a certain distance s is best to be calculated by numerical means rahter than analytical way.