標題:
FP~Maths
發問:
A missile is projected with initial speed 42ms^-1 at an angle of 30 above the horizontal. Ignoring air resistance, calculate the distance of the missile from the point of projection at the instant when it is moving downwards at an angle of 10 to the horizontal. 更新: 我唔明為甚麼有 tan 的出現！？ 可不可以解釋一下？
Take upward be positive. Initial vertical speed, uv = 42sin30* = 21 ms-1 Horizontal speed, uh = 42cos30* = 21√3 ms-1 Final vertical speed, vv = -21√3 tan10* ms-1 By vv2 = uv2 – 2gsv (-21√3 tan10*)2 = 212 – 2(10)sv sv = 20.0 m By vv = uv – gt (-21√3 tan10*) = 21 – 10t t = 2.74 s So, the horizontal distance of the missile from the starting point = 21√3 X 2.74 = 99.7 m (cor. to 3 sig. fig.) The distance from the starting point =√[99.72 + 20.02] = 102 m (cor. to 3 sig. fig.)
其他解答:
FP~Maths
發問:
A missile is projected with initial speed 42ms^-1 at an angle of 30 above the horizontal. Ignoring air resistance, calculate the distance of the missile from the point of projection at the instant when it is moving downwards at an angle of 10 to the horizontal. 更新: 我唔明為甚麼有 tan 的出現！？ 可不可以解釋一下？
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最佳解答:Take upward be positive. Initial vertical speed, uv = 42sin30* = 21 ms-1 Horizontal speed, uh = 42cos30* = 21√3 ms-1 Final vertical speed, vv = -21√3 tan10* ms-1 By vv2 = uv2 – 2gsv (-21√3 tan10*)2 = 212 – 2(10)sv sv = 20.0 m By vv = uv – gt (-21√3 tan10*) = 21 – 10t t = 2.74 s So, the horizontal distance of the missile from the starting point = 21√3 X 2.74 = 99.7 m (cor. to 3 sig. fig.) The distance from the starting point =√[99.72 + 20.02] = 102 m (cor. to 3 sig. fig.)
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