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Q.

A man approaches the foot of a tower of height $h$ units with a speed $b$ units/sec. the speed at which he approaches the vertex of the tower, when he is at a distance of $ℓ$ units from the foot, is

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a

$\frac{b}{\sqrt{h^{2} + ℓ^{2}}} units / sec$

b

$\frac{ℓ b}{\sqrt{h^{2} + ℓ^{2}}} unit / sec$

c

$\frac{b}{\sqrt{h^{2} - ℓ^{2}}} units / sec$

d

$\frac{ℓ}{b \sqrt{h^{2} + ℓ^{2}}} units / sec$

answer is B.

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Detailed Solution

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Let $y = \sqrt{l^{2} + h^{2}} & b = \frac{d l}{d t}$

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$\Rightarrow \frac{d y}{d t} = \frac{1}{2 \sqrt{h^{2} + l^{2}}} 2 l \frac{d l}{d t} = \frac{l b}{\sqrt{l^{2} + h^{2}}}$

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