f.4 物理 help

在最高點的速度 = 0


當 u1 = u, s1 = h, a1 = a, v1 = 0, t1 = t

v12 = u12 + 2a1s1

(0)2 = u2 + 2ah

u2 = -2ah ...... (1)


v1 = u1 + a1t1

0 = u + at

u = -at ...... (2)

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當 u2 = 2u, a2 = a, v2 = 0, s2 = ?, t2 = ?

v22 = u22 + 2a2s2

0 = (2u)2 + 2as2

4u2 + 2as2 = 0 ...... (3)

把 (1) 代入 (3):

4(-2ah) + 2as2 = 0

2as2 - 8ah = 0

2a(s2 - 4h) = 0

a ≠ 0，所以 s2 = 4h


v2 = u2 + a2t2

0 = (2u) + at2

2u + at2 = 0 ...... (4)

把 (2) 代入 (4):

2(-at) + at2 = 0

at2 = 2at

t2 = 2t

答: 最高點的高度 = 4h
答: 時間 = 2t
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