physic 難題 5

2011-12-30 8:29 pm

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請詳細說明, 謝謝!!

更新1:

(a) 怎樣得來[v.sin(5)]^2 = [u.sin(8)]^2 + 2.(-10).(2.15-2)??, 以我的物理書資料的公式是 v^2 = u^2 - 2g x delt height?? 另外我無論怎樣將兩個EQUATION結合, 答案與你的不同? 請指教~

回答 (1)

天同

2011-12-30 10:56 pm

✔ 最佳答案

(a) Consider the horizontal direction: u.cos(8) = v.cos(5)Consider the vertical direction: [v.sin(5)]^2 = [u.sin(8)]^2 + 2.(-10).(2.15-2)Sove the two equations for u and v gives u = 16.16 m/s and v = 16.06 m/s (b) Using equation of motion: v = u + at-16.06.sin(5) = 16.16.sin(8) + (-10).tsolve for t gives t = 0.365 shence, s = [16.16.cos(8)] x 0.365 m = 5.84 m ( c) t = 0.365 s (see part (b)) (d) Change of momentum = 0.5 x 16.16 Ns = 8.08 Ns

2011-12-31 17:28:39 補充：
Initial vertical velocity = u.sin(8)
Final vertical vel = -v.sin(5)
Hence, [-v.sin(5)]^2 = [u.sin(8)]^2 + 2.(-10).(2-15 - 2)

2011-12-31 17:29:34 補充：
It is only a matter of mathematics to solve the equations. The difference in the solution is probably due to truncation error.

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