A particle is being projected with speed u from the top of a cliff of height h above ground.
(a) If the angle of projection is @, show that the horizontal range R is given by
R = u^2 sin@cos@/g + u^2 cos@[ sin^2 @ + (2gh/u^2) ] ^0.5 /g
(b) Show that the maximum value for R is
u^2/g * [ 1 + (2gh/u^2) ]^0.5 ,
the particle being projected at an angle of
arc cot [ 1 + (2gh/u^2) ] ^0.5 .
Applied Maths - Mechanics 34
2009-08-10 11:24 pm
回答 (1)
2009-08-11 7:06 am
✔ 最佳答案
The equation of motion are:x = ucosθt … (1)
y = usinθt – gt2/2 … (2)
The range is determined by y = -h
So – h = usinθt – gt2/2
(g/2)t2 – usinθt – h = 0
t = [usinθ +/- √(u2sin2θ + 2gh)]/g
The negative value is dropped leaving
t = (u/g)[sinθ + √(sin2θ + 2gh/u2)]
Sub into (1), the range is
R = u2sinθcosθ/g + u2cosθ√(sin2θ + 2gh/u2)/g
R = (u2/g)[sinθcosθ + cosθ√(sin2θ + 2gh/u2)]
dR/dθ = (u2/g)[cosθcosθ – sinθsinθ – sinθ√(sin2θ + 2gh/u2) + cosθ(sinθcosθ)/√(sin2θ + 2gh/u2)]
Set dR/dθ = 0,
cosθcosθ – sinθsinθ – sinθ√(sin2θ + 2gh/u2) + cosθ(sinθcosθ)/√(sin2θ + 2gh/u2) = 0
Let Z = √(sin2θ + 2gh/u2) where Z > 0
cos2θ – sin2θ – Zsinθ + sinθcos2θ/Z = 0
(sinθZ – cos2θ)(Z + sinθ) = 0
Z = cos2θ/sinθ or Z = – sinθ (rejected)
Z = cos2θ/sinθ = √(sin2θ + 2gh/u2)
cos4θ/sin2θ = sin2θ + 2gh/u2
cot2θ – 1 = 2gh/u2
cotθ = √(1 + 2gh/u2)
R = (u2/g)[sinθcosθ + cosθ√(sin2θ + 2gh/u2)]
R = (u2/g)[sinθcosθ + cosθ√(sin2θ + cot2θ – 1)]
R = (u2/g)[sinθcosθ + √(sin2θcos2θ + cos4θ/sin2θ – cos2θ)]
R = (u2/g){sinθcosθ + √[(sin4θcos2θ + cos4θ – sin2cos2θ)/sin2θ]}
R = (u2/g){sinθcosθ + √[(-sin2θcos4θ + cos4θ)/sin2θ]}
R = (u2/g)[sinθcosθ + √(cos6θ/sin2θ)]
R = (u2/g)(sinθcosθ + cos3θ/sinθ)
R = (u2/g)(sin2θcosθ + cos3θ)/sinθ
R = (u2/g)(cosθ/sinθ)
R = (u2/g)√(1 + 2gh/u2)
收錄日期: 2021-04-24 10:10:01
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20090810000051KK01211