✔ 最佳答案
1.
答案是: (D) 109.7
煙囪高度 = H m
全程需時 = T s
設全程需時 T 秒。
落地前 2 秒即由 (T- 2) 秒至 T 秒。
由 (T - 2) 秒至 T 秒的位移 = (6/9)H = 2H/3
由 0 至 (T- 2) 秒的位移 = H - (2H/3) = H/3
考慮由 0 至 (T- 2) 秒:
h = (1/2)gt²
H/3 = (1/2)*9.8*(T - 2)²
H = 3*(1/2)*9.8*(T - 2)² ...... [1]
考慮由 0 至 T 秒:
h = (1/2)gt²
H = (1/2)*9.8*T²
H = (1/2)*9.8*T² ...... [2]
[1] = [2]:
3*(1/2)*9.8*(T - 2)² = (1/2)*9.8*T²
3(T - 2)² = T²
3T² - 12T + 12 = T²
T² - 6T + 6 = 0
T = [6 ± √(6² -4*6)]/2
T = 3 + √3 或 T = 3 - √3 (不合)
把 T = 3 + √3 代入[2] 中:
H = (1/2)*9.8*(3 + √3)²
H = 109.7
煙囪高度 = 109.7 m
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2.
三秒內:t = 0 至 t= 3
第三秒內:t = 2 至 t= 3
三秒內(t = 0 至 t= 3):
h = (1/2)gt²
h = (1/2)g(3)²
h = 4.5g
二秒內(t = 0 至 t= 2):
h = (1/2)gt²
h = (1/2)g(2)²
h = 2g
第三秒內位移 : 三秒內位移
= (4.5g - 2g) : 4.5g
= 2.5 : 4.5
= 5 : 9
2014-07-03 22:59:31 補充:
因為 √3 = 1.7321
H = (1/2)*9.8*(3 + √3)²
H = (1/2)*9.8*(3 + 1.7321)²
H = 109.7