在一懸崖以速率為 15ms^-1 垂直向上拋擲一物體。
10s 後物體到達地面。懸崖離地面多高 ?
位置和移動（運動方程）
2008-08-31 7:23 am
回答 (2)
2008-08-31 7:35 am
✔ 最佳答案
設懸崖離地面 s m,物體在懸崖之上的時間(2t),
v = u +at
0 = 15 + (-10)t
t = 1.5
物體在懸崖之上的時間 = 2 x 1.5 = 3s
物體在懸崖之下的時間 = 10-3 = 7s
s = ut + 0.5at^2
s = 15(7) + 0.5(10)7^2
s = 350m
所以懸崖離地面350m 高
2008-08-31 7:38 am
Using equation of motion,
u = 15 m/s, a = g = -10 m/s2, t = 10 s, s = ?
use s = ut + (1/2).a.t^2
hence, s = 15x10 +(1/2)(-10)x(10x10)
i.e. s = -350 m
The height of the cliff is 350 m
u = 15 m/s, a = g = -10 m/s2, t = 10 s, s = ?
use s = ut + (1/2).a.t^2
hence, s = 15x10 +(1/2)(-10)x(10x10)
i.e. s = -350 m
The height of the cliff is 350 m
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