物理2題....plshelp!

1) 取 g = 10 m/s2, 則:

設球在落地前一秒的瞬時速度為 u, 代入公式:

ut + at2/2 = s

u + 5 = 45

u = 40 m/s

所以球在落地前一秒的瞬時速度為 40 m/s, 即此時它已下落了 4 秒.

換言之, 它共用了 5 秒由高度 h 下落, 再代入公式:

ut + at2/2 = s, 其中 u = 0, a = 10 和 t = 5

s = 125

即 h = 125 m



2) 小球相對人的初速為垂直, 假設為 u

則球在空中逗留了 125/25 = 5 s (考慮橫向分量)

所以, 再次考慮垂直運動時, 代入1:

ut + at2/2 = s, 其中 s = 0, a = -10 和 t = 5

5u - 125 = 0

u = 25 m/s

即小球相對人的初速為向上 25 m/s.

忽略空氣阻力, 在回落拋出點的水平時, 小球相對人的初速為向下 25 m/s.

所以球著地時的瞬時速度為垂直 25 m/s 加橫向 25 m/s, 即 25√2 = 35.4 m/s.

1. Let T be the time of fall through the height h
Use equation of motion: s = ut + (1/2)a.t^2
with s = h, u = 0 m/s, t = T, a = g (=10 m/s2)
hence, h = (1/2)(10).T^2
i.e. h = 5T^2 ------------------------ (1)

Since the ball required (T-1) s to fall through a height of (h-45) m, use the equation: s = ut + (1/2)at^2 again,
with s = (h-45), u = 0 m/s, a = g(= 10 m/s2), t = (T-1)
hence, h-45 = (1/2)(10)(T-1)^2
i.e. h = 5(T-1)^2 + 45 -------------------- (2)

Equating (1) and (2):
5T^2 = 5(T-1)^2 + 45
5T^2 = 5T^2 - 10T + 5 + 45
T = 50/10 s = 5 s
Substitute T = 5s into (1) gives h = 5 x 5^2 m = 125 m

2. 豎直平面內,以Vo=25m/s的速度人準 拋出一個小球...
Don't know what the meaning of the words "人準" is. Is there a diagram associated with this problem?