物理題 請幫手解答

Let h be the height of the tower,
t be the time needed for the object to reach the ground
using the equation of motion: s = ut + (1/2)at^2
u = 0 m/s, a = g, s = h, time = t
h = (1/2)g.t^2 ----------------------- (1)
Since the object travels (9/25)h in the last second, that is to say, it travels (1-9/25).h in (t-1) seconds
using the equation s = ut + (1/2)at^2 again
u = 0 m/s, a = g, time = t-1, s = (1-9/25)h = 16h/25
16h/25 = (1/2).g.(t-1)^2 ------------------ (2)
Dividing: (2)/(1) gives
16/25 = (t-1)^2/t^2
simplifying, 9t^2 - 50t +25 = 0
solve for t gives t = 5/8 s (rejected because t must be longer than 1 s)
or t = 5 s (accepted)
substitute t = 5s into (1), and taking g = 10 m/s2
h = (10 x 5 x 5)/2 m = 125 m

設物体到达地面的velocity是x

到达地面前最后一秒開始的velocity是(x-10)

v(二次) - u(二次)=2as

(x)二次 -(x-10)二次=2(10)(9／25)s

x=(7.2s+100)／20



u在塔顶是0

x=(7.2s+100)／20........................1

v(二次) - u(二次)=2as..........2

put 1 into 2

0.1296s-16.4s+25=0

s=(16.4+／- 16)／0.2592

s=125m or1.543209877(rej)

2009-05-18 20:37:49 補充：
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2009-05-18 20:39:50 補充：
地心+ x度(accleration due to gravity) 是10 m/s2