Find α in terms of θ using the equations: (i) u^2 sin^2 α = 2gd cos θ (ii) t = ((u cos α)/(g sin θ)) (iii) −d = ut sin α − ((gt^2 sin θ)/2)


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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
	$Substituting the value of t in (iii) −d = u(((u cos α)/(g sin θ)))sin α − g(((u cos α)/(g sin θ)))^2 ((sin)/2) −dgsin θ=u^2 cos αsin α−(1/2)u^2 cos^2 α −2dgsinθ=u^2 sin2α−u^2 cos^2 α −2dgsinθ=u^2 sin2α−u^2 (1−sin^2 α) u^2 −2dgsinθ=u^2 sin2α+u^2 sin^2 α u^2 −2dgsinθ=u^2 sin2α+2dgcosθ u^2 −2dgsinθ−2dgcosθ =u^2 sin2α u^2 sin2α=u^2 −2dg(sinθ+cosθ) sin2α=1−((2dg(sinθ+cosθ))/u^2 ) α=(1/2)sin^(−1) {1−((2dg(sinθ+cosθ))/u^2 ) }$
	$S u b s t i t u t i n g t h e v a l u e o f t i n (i i i)$ $- d = u (\frac{u \cos α}{g \sin θ}) \sin α - g {(\frac{u \cos α}{g \sin θ})}^{2} \frac{\sin}{2}$ $- d g \sin θ = u^{2} \cos α \sin α - \frac{1}{2} u^{2} \cos^{2} α$ $- 2 d g \sin θ = u^{2} \sin 2 α - u^{2} \cos^{2} α$ $- 2 d g \sin θ = u^{2} \sin 2 α - u^{2} (1 - \sin^{2} α)$ $u^{2} - 2 d g \sin θ = u^{2} \sin 2 α + u^{2} \sin^{2} α$ $u^{2} - 2 d g \sin θ = u^{2} \sin 2 α + 2 d g \cos θ$ $u^{2} - 2 d g \sin θ - 2 d g \cos θ = u^{2} \sin 2 α$ $u^{2} \sin 2 α = u^{2} - 2 d g (\sin θ + \cos θ)$ $\sin 2 α = 1 - \frac{2 d g (\sin θ + \cos θ)}{u^{2}}$ $α = \frac{1}{2} \sin^{- 1} {1 - \frac{2 d g (\sin θ + \cos θ)}{u^{2}}}$
	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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