An arch way is in the shapeof a semi ellipse; the road level being the major axis. If the breadth of the road is $30 ⨍ t$ and a man $6 ⨍ t$ tall just touches the top when $2 ⨍ t$ from the side then the greatest height of the arch is

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$\frac{45}{\sqrt{14}} ft$

$\frac{15}{\sqrt{14}} ft$

$\frac{90}{\sqrt{221}} ft$

$\frac{\sqrt{14}}{3} ft$

answer is A.

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Detailed Solution

$L e t A B A' b e t h e s e m i e l l i p t i c a r c h s u c h t h a t A A' = 30 a n d l e t O B = b l e t t h e e q u a t i o n o f e l l i p s e b e \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 L e t P (x, y) b e a n y p o i n t a t a d i s \tan c e 2 f e e t f r o m A ∴ O P = O A - P A = 13 A A' = 2 a = 30 \Rightarrow a = 15 t h e p o i n t Q (13, 6) l i e s o n e l l i p o s e t h e n \frac{169}{225} + \frac{36}{b^{2}} = 1 \Rightarrow \frac{36}{b^{2}} = \frac{56}{225} \Rightarrow b^{2} = \frac{36 \times 225}{56} \Rightarrow b = \sqrt{\frac{225 \times 9}{14}} = \frac{45}{\sqrt{14}}$