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Question Number 13705 by ajfour last updated on 22/May/17

A s s u m i n g n o a i r r e s i s t a n c e,

a n d a n g l e o f p r o j e c t i o n α = \frac{π}{4},

f i n d t h e r a t i o o f t h e l e n g t h o f

t r a j e c t o r y L o f a p r o j e c t i l e m o t i o n

(b y t h e t i m e i t h i t s t h e g r o u n d)

t o i t s h o r i z o n t a l r a n g e R o n

g r o u n d . \frac{L}{R} = ?

Commented by ajfour last updated on 22/May/17

Commented by mrW1 last updated on 22/May/17

f o r p a r a b o l a f (x) = {a x}^{2}

t h e l e n g t h o f t h e c u r v e f r o m x = 0 t o

x = b i s

l = \int_{0}^{b} \sqrt{1 + {(\frac{d y}{d x})}^{2}} d x

= \int_{0}^{b} \sqrt{1 + {(2 a x)}^{2}} d x

= \frac{1}{2 a} \int_{0}^{2 a b} \sqrt{1 + t^{2}} d t

= \frac{1}{4 a} {[t \sqrt{1 + t^{2}} + \ln (t + \sqrt{1 + t^{2}})]}_{0}^{2 a b}

= \frac{b}{2} \times \frac{1}{2 a b} (2 a b \sqrt{1 + {(2 a b)}^{2}} + \ln \sqrt{1 + {(2 a b)}^{2}})

a p p l y t h i s i n o u r c a s e :

b = \frac{R}{2}

h e i g h t o f p a r a b o l a h i s

h = \frac{R}{4} \times \tan α_{0} = \frac{R}{4} \times \tan \frac{π}{4} = \frac{R}{4}

h = a {(\frac{R}{2})}^{2} = a \frac{R^{2}}{4} = \frac{R}{4}

\Rightarrow a = \frac{1}{R}

2 a b = 2 \times \frac{1}{R} \times \frac{R}{2} = 1

\Rightarrow L = 2 \times \frac{R}{4} \times \frac{1}{1} \times [1 \times \sqrt{1 + 1^{2}} + \ln (1 + \sqrt{1 + 1^{2}})] = \frac{R}{2} \times [\sqrt{2} + \ln (1 + \sqrt{2})]

\Rightarrow \frac{L}{R} = \frac{\sqrt{2} + \ln (1 + \sqrt{2})}{2} \approx 1.15

Commented by ajfour last updated on 22/May/17

t h a n k y o u S i r, o u r a n s w e r s e v e n

m a t c h!

Answered by ajfour last updated on 22/May/17

d L = \sqrt{1 + {(\frac{d y}{d x})}^{2}} d x

d y = (u \sin α - g t) d t

d x = (u \cos α) d t

\frac{d y}{d x} = (\frac{u \sin α - g t}{u \cos α})

\int_{0}^{L} d L = \int_{0}^{T} \sqrt{1 + {(\frac{u \sin α - g t}{u \cos α})}^{2}} (u \cos α) d t

f o r y c o m p o n e n t o f m o t i o n

v_{y} = u \sin α - g t

= 0 f o r t = \frac{T}{2}

\Rightarrow T = \frac{2 u \sin α}{g}

f o r α = \frac{π}{4}, T = \frac{u \sqrt{2}}{g} \dots . . (i)

L = \frac{u}{\sqrt{2}} \int_{0}^{u \sqrt{2} / g} \sqrt{1 + {(1 - \frac{g t \sqrt{2}}{u})}^{2}} d t

l e t p = 1 - \frac{g t \sqrt{2}}{u}

d p = - \frac{g \sqrt{2}}{u} d t \Rightarrow d t = - \frac{u}{g \sqrt{2}} d p

w h e n t = 0, p = 1

w h e n t = \frac{u \sqrt{2}}{g}, p = - 1

∴ L = \frac{u}{\sqrt{2}} \int_{1}^{- 1} \sqrt{1 + p^{2}} (- \frac{u}{g \sqrt{2}}) d p

L = \frac{u^{2}}{g} \int_{0}^{1} \sqrt{1 + p^{2}} d p

L = \frac{u^{2}}{g} {[\frac{p}{2} \sqrt{1 + p^{2}} + \frac{1}{2} \ln (p + \sqrt{1 + p^{2}})]}_{0}^{1}

L = \frac{u^{2}}{2 g} [\sqrt{2} + \ln (1 + \sqrt{2})]

R = (u \cos α) T

= \frac{u}{\sqrt{2}} \frac{u \sqrt{2}}{g} = \frac{u^{2}}{g}

s o, \frac{L}{R} = \frac{\sqrt{2} + \ln (1 + \sqrt{2})}{2} \approx 1.148

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