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Question Number 46482 by peter frank last updated on 27/Oct/18

Commented by peter frank last updated on 27/Oct/18

$help$
	Answered by tanmay.chaudhury50@gmail.com last updated on 28/Oct/18

	$h o r i z o n t a l v e l o c i t y o f p r o j e c t i o n = v c o s α$ $v e r t i c a l v e l o c i t y o f p r o j e c t i o n = v s i n α$ $e q n f o r h o r i g o n t a l m o t i o n$ $x = (v c o s α) t$ $e q n f o r v d r t i c a l m o t i o n$ $y = (v s i n α) t - \frac{1}{2} {g t}^{2}$ $t h e r e i s n o a c c e l a r t i o n a l o n g h o r i z o n t a l d i r e c t i o n$ $a c c e l a r a t i o n (g) i s a c t i n g v e r t i c a l l y d o w n$ $a f t e r t i m e t_{0} t h e v e r t i c a l v e l o c i t y = 0$ $0 = v s i n α - {g t}_{0} t_{0} = \frac{v s i n α}{g}$ $t o t a l t i m e o f f l i g h t = 2 t_{0}$ $= \frac{2 v s i n α}{g}$ $R a n g e = (v c o s α) (\frac{2 v s i n α}{g}) = v^{2} \times \frac{2 s i n α c o s α}{g} = \frac{v^{2} s i n 2 α}{g}$ $g r e a t e s t h e i g h t w h e n v e r t i c a l c o m p o n e n t o f$ $v e l o c i t y i s z e r o .$ $0^{2} = {(v s i n α)}^{2} - 2 g H [f o r m u l a v^{2} = u^{2} - 2 a s]$ $H = \frac{v^{2} {s i n}^{2} α}{2 g}$ $y = (v s i n α) t - \frac{1}{2} {g t}^{2}$ $y = (v s i n α) (\frac{x}{v c o s α}) - \frac{1}{2} g {(\frac{x}{v c o s α})}^{2}$ $y = x t a n α - \frac{1}{2} g (\frac{x^{2}}{v^{2} {c o s}^{2} α})$ $y = A x - {B x}^{2}$ ${B x}^{2} - A x = - y$ $x^{2} - \frac{A}{B} x = - \frac{y}{B}$ $x^{2} - 2 . x . \frac{A}{2 B} + \frac{A^{2}}{4 B^{2}} = - \frac{y}{B} + \frac{A^{2}}{4 B^{2}}$ ${(x - \frac{A}{2 B})}^{2} = (\frac{A^{2} - 4 y B}{4 B^{2}})$ ${(x - \frac{A}{2 B})}^{2} = (- 4 B) \times \frac{1}{4 B^{2}} (y - \frac{A^{2}}{4 B})$ ${(x - \frac{A}{2 B})}^{2} = (- 4) \times (\frac{1}{4 B}) (y - \frac{A^{2}}{4 B})$ $X^{2} = - 4 P Y$ $f o c u s (0, - P)$ $x - \frac{A}{2 B} = 0 x = \frac{A}{2 B}$ $y - \frac{A^{2}}{4 B} = - P$ $y - \frac{A^{2}}{4 B} = - \frac{1}{4 B}$ $y = \frac{A^{2} - 1}{4 B}$ $f o c u s (\frac{A}{2 B}, \frac{A^{2} - 1}{4 B})$ $d i r e c t r i x Y = P$ $y - \frac{A^{2}}{4 B} = \frac{1}{4 B}$ $y = \frac{A^{2} + 1}{4 B}$
	Commented by peter frank last updated on 28/Oct/18

	$how do i get focus and directrix$
	Commented by peter frank last updated on 28/Oct/18

	$God bless you$
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