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物理2題....plshelp!
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最佳解答:
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.
其他解答:
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?
物理2題....plshelp!
發問:
此文章來自奇摩知識+如有不便請留言告知
(1)一球自某一高度h處自由下落,已知它在落地前的一秒內所下落的高度為45米,求原來的高度h. (2)豎直平面內,以Vo=25m/s的速度人準拋出一個小球,若小球著地與拋出點的水平距離為125m,求小球著地時的瞬時速度V的大小? "木人計得55.9m/s,但答案為35.4m/s....請高手講解..... thanks!!!最佳解答:
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.
其他解答:
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?
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