A man of 5 text{ feet} height is walking away | vedclass

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(B) Let $H = 15 	ext{ ft}$ be the height of the light and $h = 5 	ext{ ft}$ be the height of the man.Let $x$ be the distance of the man from the lig

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) Let $H = 15 	ext{ ft}$ be the height of the light and $h = 5 	ext{ ft}$ be the height of the man.Let $x$ be the distance of the man from the light source and $s$ be the length of his shadow.By similar triangles, we have $\frac{s}{h} = \frac{x+s}{H}$.Substituting the values, $\frac{s}{5} = \frac{x+s}{15} \implies 3s = x + s \implies 2s = x$.Differentiating with respect to time $t$, we get $2 \frac{ds}{dt} = \frac{dx}{dt}$.Given $\frac{ds}{dt} = \frac{11}{5} 	ext{ ft/sec}$, so $\frac{dx}{dt} = 2 	imes \frac{11}{5} = \frac{22}{5} 	ext{ ft/sec}$.To convert $\frac{dx}{dt}$ to $	ext{miles/hour}$, we have $\frac{22}{5} 	ext{ ft/sec} = \frac{22}{5} 	imes 3600 	ext{ ft/hour} = \frac{22 	imes 3600}{5 	imes 5280} 	ext{ miles/hour}$.Calculating this, $\frac{79200}{26400} = 3 	ext{ miles/hour}$.Thus, $K = 3$.

$A$ man of $5 \text{ feet}$ height is walking away from a light fixed at a height of $15 \text{ feet}$ at the rate of $K \text{ miles/hour}$. If the rate of increase of his shadow is $\frac{11}{5} \text{ feet/sec}$, then $K=$ (Take $1 \text{ mile} = 5280 \text{ feet}$)

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