Math Question and Answers


Question and Answers Forum

All Questions Topic List
Geometry Questions
Previous in All Question Next in All Question
Previous in Geometry Next in Geometry
Question Number 33452 by ajfour last updated on 16/Apr/18

Commented by ajfour last updated on 16/Apr/18

$F i n d l e n g t h o f s h a d o w s o f a$ $s t i c k o f l e n g t h l w h e n s o u r c e$ $o f l i g h t i s a p o i n t s o u r c e a t$ $h e i g h t H a b o v e g r o u n d .$
Commented by ajfour last updated on 19/Apr/18

$q u e s t i o n f o r y o u, m r W S i r . . .$
	Answered by MrW3 last updated on 12/Jun/18

	Commented by MrW3 last updated on 12/Jun/18

	$x_{1} = 0, y_{1} = H$ $x_{2} = u, y_{2} = v$ $x_{3} = u + l \cos α, y_{3} = v + l \sin α$ $e q n . o f l i n e 12 :$ $\frac{x - 0}{y - H} = \frac{u - 0}{v - H}$ $\Rightarrow x = \frac{u}{v - H} (y - H)$ $p o i n t 4 :$ $x_{4} = \frac{u}{v - H} (0 - H) = \frac{u H}{H - v}$ $e q n . o f l i n e 13 :$ $\frac{x - 0}{y - H} = \frac{u + l \cos α - 0}{v + l \sin α - H}$ $\Rightarrow x = \frac{u + l \cos α}{v + l \sin α - H} (y - H)$ $p o i n t 5 :$ $x_{5} = \frac{u + l \cos α}{v + l \sin α - H} (0 - H) = \frac{(u + l \cos α) H}{H - v - l \sin α}$ $s = x_{5} - x_{4} = \frac{(u + l \cos α) H}{H - v - l \sin α} - \frac{u H}{H - v}$ $= \frac{(u + l \cos α) (H - v) - u (H - v - l \sin α)}{(H - v - l \sin α) (H - v)} \times H$ $\Rightarrow s = \frac{(H - v) \cos α + u \sin α}{(H - v - l \sin α) (H - v)} \times l H$ $o r$ $\Rightarrow s = \frac{(H - y) \cos α + x \sin α}{(H - y - l \sin α) (H - y)} \times l H$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum