A man of height 2 text{ m} walks at a uniform | vedclass

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Question

(A) Let $AB$ be the lamp post of height $6 	ext{ m}$ and $MN$ be the man of height $2 	ext{ m}$. Let the man be at a distance $l$ from the lamp post

Vedclass · Accepted Answer

$2.5$

Vedclass · Answer

(A) Let $AB$ be the lamp post of height $6 	ext{ m}$ and $MN$ be the man of height $2 	ext{ m}$. Let the man be at a distance $l$ from the lamp post at time $t$,so $AM = l$. Let $MS = s$ be the length of the shadow.Since $	riangle MSN \sim 	riangle ASB$,we have:$\frac{MS}{AS} = \frac{MN}{AB}$Substituting the values,we get:$\frac{s}{l + s} = \frac{2}{6}$$\frac{s}{l + s} = \frac{1}{3}$$3s = l + s$$l = 2s$Differentiating both sides with respect to time $t$:$\frac{dl}{dt} = 2 \frac{ds}{dt}$Given that the man walks at a speed of $\frac{dl}{dt} = 5 	ext{ km/h}$,we have:$5 = 2 \frac{ds}{dt}$$\frac{ds}{dt} = \frac{5}{2} = 2.5 	ext{ km/h}$Thus,the length of the shadow increases at the rate of $2.5 	ext{ km/h}$.

$A$ man of height $2 \text{ m}$ walks at a uniform speed of $5 \text{ km/h}$ away from a lamp post which is $6 \text{ m}$ high. Find the rate at which the length of his shadow increases. (in $\text{ km/h}$)

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