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Question Number 21441 by Tinkutara last updated on 23/Sep/17

Find α in terms of θ using the equations :

(i) u^{2} \sin^{2} α = 2 g d \cos θ

(i i) t = \frac{u \cos α}{g \sin θ}

(i i i) - d = u t \sin α - \frac{{g t}^{2} \sin θ}{2}

Answered by $@ty@m last updated on 24/Sep/17

S u b s t i t u t i n g t h e v a l u e o f t i n (i i i)

- d = u (\frac{u \cos α}{g \sin θ}) \sin α - g {(\frac{u \cos α}{g \sin θ})}^{2} \frac{\sin}{2}

- d g \sin θ = u^{2} \cos α \sin α - \frac{1}{2} u^{2} \cos^{2} α

- 2 d g \sin θ = u^{2} \sin 2 α - u^{2} \cos^{2} α

- 2 d g \sin θ = u^{2} \sin 2 α - u^{2} (1 - \sin^{2} α)

u^{2} - 2 d g \sin θ = u^{2} \sin 2 α + u^{2} \sin^{2} α

u^{2} - 2 d g \sin θ = u^{2} \sin 2 α + 2 d g \cos θ

u^{2} - 2 d g \sin θ - 2 d g \cos θ = u^{2} \sin 2 α

u^{2} \sin 2 α = u^{2} - 2 d g (\sin θ + \cos θ)

\sin 2 α = 1 - \frac{2 d g (\sin θ + \cos θ)}{u^{2}}

α = \frac{1}{2} \sin^{- 1} {1 - \frac{2 d g (\sin θ + \cos θ)}{u^{2}}}

Commented by Tinkutara last updated on 24/Sep/17

What if the 3^{rd} equation was

- d = u t \sin α - \frac{{g t}^{2} \cos θ}{2} ?

Commented by $@ty@m last updated on 24/Sep/17

i n t h a t c a s e t w o u l d b e \frac{u \cos α}{g \cos θ}

a n d t h e p r o c e d u r e w o u l d b e

s i m i l a r .

Commented by Tinkutara last updated on 24/Sep/17

Not everywhere \cos θ, only in 3^{rd}

equation .

Commented by $@ty@m last updated on 24/Sep/17

I n t h a t c a s e i t w o u l d r e s u l t i n

a n i m p l i c i t f u n c t i o n .

Commented by Tinkutara last updated on 24/Sep/17

Yes, can you solve it ? Because it is

required in a Physics question .

Commented by $@ty@m last updated on 24/Sep/17

I n a n i m p l i c i t f n c t i o n

o n e v a r i a b l e c a n n o t b e e x p r e s s e d

i n t e r m s o f a n o t h e r v a r i a b l e .

I n m y o p n i o n,

i n s u c h P h y s i c s q u e s t i o n,

t h e r e w o u l d b e o n e m o r e r e l a t i o n

b e t w e e n t h e v a r i a b l e s .

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