物理之力學問題

To solve this problem you need to do two things:

(1)find the velocity and position of the first ball after 4s.
(2)using the position they meet (same displacement from the point
they got thrown) to set an equation to find the time they meet, then
find the position.

Assuming upward direction of y-coordinate to be position, and point A as the origin.

(1)find the velocity and position

V = Vo + at
==> V = 24 + (-9.8)(4)
==> V = -15.2 m/s


S = So +Vot +1/2(at^2)
==> S = 0 + 24(4) + 1/2 (-9.8)(4^2)
==> S = 17.6 m

(2)find the time they meet

[So + Vot +(1/2)at^2] (for ball 1) = [So + Vot +(1/2)at^2] (for ball 2)
==> 17.6 + (-15.2)t + (1/2)(-9.8)t^2 = 0 + (-10)t +(1/2)(-9.8)(t^2)
==> 17.6 −15.2t - (1/2)(9.8)t^2 = −10t - (1/2)(9.8)t^2
==> 17.6 = 5.2t
==> t = 3.38 s

Hence they will meet at

S = So + Vot + (1/2)a(t^2) [for ball 2]
==> S = 0 + (-10)(3.38) + (1/2)(-9.8)(3.38^2)
==> S = -89.78 m <— answer (89.78 below point A)



2012-12-05 23:01:01 補充：
in the question statement, it has "B is 5 m below point A". I don't know why it has such statement. If the balls can only travel 5m below point A, then there is NO need to find the position they meet.

2012-12-06 02:55:40 補充：
If you use S = So + Vot + (1/2)a(t^2) [for ball 1], you will a little different answer, because truncation error.

2012-12-06 08:34:28 補充：
Oh! I think, this question might be just part of the question.

2012-12-12 03:52:50 補充：
Haven't done this for a long while! Pretty soon, it will give me headache too!

MSG is also good @ equations I see....

This question is giving me a major headache....I am outta here!

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