maths..F.4

b) s=10t-5t^2
s= -5(t^2-2t)
s= -5(t^2-2t+1)+5
s= -5(t-1)^2 +5
As maximum of -5(t-1)^2=0
So that maximum of s = 5

a)maximum of s = 5
10t-5t^2=5
5t^2-10t+5=0
t^2-2t+1=0
(t-2)(t+1)=0
t=2 or t=-1(rejected)

So that the bullet attent its maximum height 5 unit when t =2unit

2006-10-25 22:44:13 補充：
sorry, It's wrong.t^2-2t+1=0(t-1)^2=0t=1(repeated)So that the bullet attent its maximum height 5 unit when t =1 unit.

s= 10t-5t^2
s=-5(t^2-2t)
s=-5[t^2-2t+(1)^2-(1)^2]
s=-5[(t-1)^2-1]
s=-5(t-1)+5

a)when the time is 1(你冇俾單位) ,
the bullet will reach the maximum height

b)the maximum height is 5m that the bullet can reach

因為你既t 係個time , 當t係1既時候就會到最大值 ...
所以當t係1就會到the maximum height

而當t 係1既時候 , 個s 就會係5 ...
所以個 maximum height 就係5m

s = 10 t - 5 t^2
s = - 5 t^2 + 10 t
s = - 5 ( t ^ 2 - 2 t )
s = - 5 [ t ^ 2 - 2 t + (1) ^ 2 ] - (-5) (1) ^ 2
s = - 5 ( t - 1 ) ^ 2 + 5

So, bullet reaches the maximum height 5 when t = 1