phy 問題..高手請進

Take g = 10 m s-2
Take upward quantities to be positive.

(a)
考慮由跳板至最高點的運動，所需時間是 0.7 s。
以向上速度去到最高點後，改以向下速度下墜。所以最高點的速度是 0 m s-1。
加速度是向下的，所以是負數，即 -10 m s-2。(即 -g，依題目所給的 g 值)
t = 0.7 s, a = -10 m s-2, v = 0 m s-1
v = u + at
0 = u + (-10)(0.7)
u = 7 m s-1
Speed when she leaves the spring board = 7 m s-1

(b)
考慮由彈板至水面的整段旅程。起始速度在 (a) 部份已計算。而加速度也沒有改變。
所需時間是向上至最高點的0.7秒，與向下一段所需的1.5秒之和。
u = 7 m s-1, t = 0.7+1.5 = 2.2 s, a = -10 m s-2
s = ut + (1/2)at2
s = 7(2.2) + (1/2)(-10)(2.2)2
s = -8.8 m
Height of the spring board above the water = 8.8 m

(c)
考慮情況與 (b) 部份一樣，不過使用不同的公式去計算末速。
u = 7 m s-1, t = 0.7+1.5 = 2.2 s, a = -10 m s-2
v = u + at
v = (7) + (-10)(2.2)
v = -15 m s-1
Speed when she enters the water = 1.5 m s-1
=

2008-12-27 01:18:49 補充：
The last line should be:
Speed when she enters the water = 15 m s^-1